Let a line L$L$ passing through the point P(1, 1, 1)$P(1, 1, 1)$ be perpendicular to the lines fracx - 44 = fracy - 11 = fracz - 11$\frac{x - 4}{4} = \frac{y - 1}{1} = \frac{z - 1}{1}$ and fracx - 171 = fracy - 711 = fracz0$\frac{x - 17}{1} = \frac{y - 71}{1} = \frac{z}{0}$. Let the line L$L$ intersect the yz-plane at the point Q$Q$. Another line parallel to L$L$ and passing through the point S(1, 0, -1)$S(1, 0, -1)$ intersects the yz-plane at the point R$R$. Then the square of the area of the parallelogram PQRS$PQRS$ is equal to

### Related Formula textLength of projection of vecu text on vecw = frac|vecu cdot vecw||vecw|$$\text{Length of projection of } \vec{u} \text{ on } \vec{w} = \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|}$$ ### Core Logic Given point A(5, 4, 2)$A(5, 4, 2)$ and line (L)$(L)$: vecr = (-hati + 3hatj + hatk) + lambda(2hati + 3hatj - hatk)$$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$$ Any general point P$P$ on this line has coordinates: (-1 + 2lambda, 3 + 3lambda, 1 - lambda)$(-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$ ### Step 1: Finding the foot of the perpendicular Vector vecAP = P - A = (-1 + 2lambda - 5)hati + (3 + 3lambda - 4)hatj + (1 - lambda - 2)hatk$\vec{AP} = P - A = (-1 + 2\lambda - 5)\hat{i} + (3 + 3\lambda - 4)\hat{j} + (1 - \lambda - 2)\hat{k}$ vecAP = (2lambda - 6)hati + (3lambda - 1)hatj + (-lambda - 1)hatk$$\vec{AP} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$$ Since AP$AP$ is perpendicular to line (L)$(L)$, the dot product of vecAP$\vec{AP}$ with the direction vector of the line (2hati + 3hatj - hatk)$(2\hat{i} + 3\hat{j} - \hat{k})$ must be zero: vecAP cdot (2hati + 3hatj - hatk) = 0$$\vec{AP} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$$ 2(2lambda - 6) + 3(3lambda - 1) - 1(-lambda - 1) = 0$$2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0$$ 4lambda - 12 + 9lambda - 3 + lambda + 1 = 0$$4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0$$ 14lambda - 14 = 0 Rightarrow lambda = 1$$14\lambda - 14 = 0 \Rightarrow \lambda = 1$$

### Step 2: Coordinates of the foot Substitute lambda = 1$\lambda = 1$ into general point P$P$ to get (alpha, beta, gamma)$(\alpha, \beta, \gamma)$: alpha = -1 + 2(1) = 1$\alpha = -1 + 2(1) = 1$ beta = 3 + 3(1) = 6$\beta = 3 + 3(1) = 6$ gamma = 1 - 1 = 0$\gamma = 1 - 1 = 0$ Foot of perpendicular is (1, 6, 0)$(1, 6, 0)$. ### Step 3: Calculate the projection Let vecu = alphahati + betahatj + gammahatk = hati + 6hatj + 0hatk$\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k}$ Let vecw = 6hati + 2hatj + 3hatk$\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$ textProjection = frac|vecu cdot vecw||vecw| = frac|1(6) + 6(2) + 0(3)|sqrt6^2 + 2^2 + 3^2$$\text{Projection} = \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|} = \frac{|1(6) + 6(2) + 0(3)|}{\sqrt{6^2 + 2^2 + 3^2}}$$ = frac6 + 12sqrt36 + 4 + 9 = frac18sqrt49 = frac187$$= \frac{6 + 12}{\sqrt{36 + 4 + 9}} = \frac{18}{\sqrt{49}} = \frac{18}{7}$$ ### Pattern Recognition Foot of perpendicular problems algorithm: 1) Frame general vector P(lambda)$P(\lambda)$. 2) Construct distance vector vecAP$\vec{AP}$. 3) Dot product with direction vector vecd = 0$\vec{d} = 0$. 4) Solve for lambda$\lambda$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra

### Related Formula textEquation of a line: fracx - x_1a = fracy - y_1b = fracz - z_1c = lambda$$\text{Equation of a line: } \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$$ textDistance formula: d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$\text{Distance formula: } d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$ ### Core Logic

Line L$L$ makes equal angles with the positive axes, so its direction ratios are (1, 1, 1)$(1, 1, 1)$. Equation of line L$L$ is fracx + 31 = fracy - 51 = fracz - 21 = lambda$\frac{x + 3}{1} = \frac{y - 5}{1} = \frac{z - 2}{1} = \lambda$. A general point R$R$ on the line is (lambda - 3, lambda + 5, lambda + 2)$(\lambda - 3, \lambda + 5, \lambda + 2)$. Use the perpendicularity condition overrightarrowPR cdot vecd = 0$\overrightarrow{PR} \cdot \vec{d} = 0$ to find lambda$\lambda$, then substitute back into the distance formula. ### Step 1: Apply Perpendicularity Condition Let point P = (-2, r, 1)$P = (-2, r, 1)$. The vector overrightarrowPR = langle lambda - 1, lambda + 5 - r, lambda + 1 rangle$\overrightarrow{PR} = \langle \lambda - 1, \lambda + 5 - r, \lambda + 1 \rangle$. Since overrightarrowPR cdot langle 1, 1, 1 rangle = 0$\overrightarrow{PR} \cdot \langle 1, 1, 1 \rangle = 0$: (lambda - 1)(1) + (lambda + 5 - r)(1) + (lambda + 1)(1) = 0$$(\lambda - 1)(1) + (\lambda + 5 - r)(1) + (\lambda + 1)(1) = 0$$ 3lambda - r + 5 = 0 implies lambda = fracr - 53$$3\lambda - r + 5 = 0 \implies \lambda = \frac{r - 5}{3}$$ ### Step 2: Calculate Distance Substitute lambda$\lambda$ back into point R$R$: R equiv left( fracr - 143, fracr + 103, fracr + 13 right)$$R \equiv \left( \frac{r - 14}{3}, \frac{r + 10}{3}, \frac{r + 1}{3} \right)$$ We are given PR = sqrtfrac143 implies PR^2 = frac143$PR = \sqrt{\frac{14}{3}} \implies PR^2 = \frac{14}{3}$. PR^2 = left( fracr - 143 + 2 right)^2 + left( fracr + 103 - r right)^2 + left( fracr + 13 - 1 right)^2 = frac143$PR^2 = \left( \frac{r - 14}{3} + 2 \right)^2 + \left( \frac{r + 10}{3} - r \right)^2 + \left( \frac{r + 1}{3} - 1 \right)^2 = \frac{14}{3}$ frac(r - 8)^29 + frac(10 - 2r)^29 + frac(r - 2)^29 = frac143$$\frac{(r - 8)^2}{9} + \frac{(10 - 2r)^2}{9} + \frac{(r - 2)^2}{9} = \frac{14}{3}$$ (r^2 - 16r + 64) + (100 + 4r^2 - 40r) + (r^2 - 4r + 4) = 42$$(r^2 - 16r + 64) + (100 + 4r^2 - 40r) + (r^2 - 4r + 4) = 42$$ 6r^2 - 60r + 168 = 42 implies 6r^2 - 60r + 126 = 0$$6r^2 - 60r + 168 = 42 \implies 6r^2 - 60r + 126 = 0$$ Dividing by 6: r^2 - 10r + 21 = 0$$r^2 - 10r + 21 = 0$$ ### Step 3: Solve for r (r - 7)(r - 3) = 0 implies r = 3, 7$$(r - 7)(r - 3) = 0 \implies r = 3, 7$$ The sum of all possible values of r$r$ is 3 + 7 = 10$3 + 7 = 10$. ### Pattern Recognition When a line makes equal angles with coordinate axes, its direction cosines are (1/sqrt3, 1/sqrt3, 1/sqrt3)$(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$, making its simpler direction ratios (1, 1, 1)$(1, 1, 1)$. Use projection vector methods or direct dot product to find perpendicular foot. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textEquation of a line: vecr = veca + lambdavecb$$\text{Equation of a line: } \vec{r} = \vec{a} + \lambda\vec{b}$$ |overrightarrowCD|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$|\overrightarrow{CD}|^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$ ### Core Logic Write general points on L_1$L_1$ and L_2$L_2$ representing points C$C$ and D$D$. The vector overrightarrowCD$\overrightarrow{CD}$ must be parallel to the given direction vector of L_3$L_3$. This yields proportional equations to solve for the line parameters. ### Step 1: Write Line Equations Line L_1$L_{1}$: fracx-2-3 = fracy-62 = fracz-74 = lambda_1$\frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4} = \lambda_{1}$ Point C$C$ on L_1$L_{1}$: (-3lambda_1+2, 2lambda_1+6, 4lambda_1+7)$(-3\lambda_{1}+2, 2\lambda_{1}+6, 4\lambda_{1}+7)$ Line L_2$L_{2}$: fracx-42 = fracy-31 = fracz-53 = lambda_2$\frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3} = \lambda_{2}$ Point D$D$ on L_2$L_{2}$: (2lambda_2+4, lambda_2+3, 3lambda_2+5)$(2\lambda_{2}+4, \lambda_{2}+3, 3\lambda_{2}+5)$ ### Step 2: Proportionality of Vector CD The vector overrightarrowCD = langle 2lambda_2+3lambda_1+2, lambda_2-2lambda_1-3, 3lambda_2-4lambda_1-2 rangle$\overrightarrow{CD} = \langle 2\lambda_{2}+3\lambda_{1}+2, \lambda_{2}-2\lambda_{1}-3, 3\lambda_{2}-4\lambda_{1}-2 \rangle$. Since L_3$L_3$ is parallel to -3hati + 5hatj + 16hatk$-3\hat{i} + 5\hat{j} + 16\hat{k}$, the components are proportional: frac2lambda_2+3lambda_1+2-3 = fraclambda_2-2lambda_1-35 = frac3lambda_2-4lambda_1-216$$\frac{2\lambda_{2}+3\lambda_{1}+2}{-3} = \frac{\lambda_{2}-2\lambda_{1}-3}{5} = \frac{3\lambda_{2}-4\lambda_{1}-2}{16}$$ ### Step 3: Solve for lambda values From the first two expressions: 5(2lambda_2+3lambda_1+2) = -3(lambda_2-2lambda_1-3)$$5(2\lambda_{2}+3\lambda_{1}+2) = -3(\lambda_{2}-2\lambda_{1}-3)$$ 10lambda_2 + 15lambda_1 + 10 = -3lambda_2 + 6lambda_1 + 9$$10\lambda_{2} + 15\lambda_{1} + 10 = -3\lambda_{2} + 6\lambda_{1} + 9$$ 13lambda_2 + 9lambda_1 = -1$$13\lambda_{2} + 9\lambda_{1} = -1$$ From the last two expressions: 16(lambda_2-2lambda_1-3) = 5(3lambda_2-4lambda_1-2)$$16(\lambda_{2}-2\lambda_{1}-3) = 5(3\lambda_{2}-4\lambda_{1}-2)$$ 16lambda_2 - 32lambda_1 - 48 = 15lambda_2 - 20lambda_1 - 10$$16\lambda_{2} - 32\lambda_{1} - 48 = 15\lambda_{2} - 20\lambda_{1} - 10$$ lambda_2 - 12lambda_1 = 38$$\lambda_{2} - 12\lambda_{1} = 38$$ Substitute lambda_2 = 12lambda_1 + 38$\lambda_{2} = 12\lambda_{1} + 38$ into the first equation: 13(12lambda_1 + 38) + 9lambda_1 = -1$$13(12\lambda_{1} + 38) + 9\lambda_{1} = -1$$ 156lambda_1 + 494 + 9lambda_1 = -1$$156\lambda_{1} + 494 + 9\lambda_{1} = -1$$ 165lambda_1 = -495 implies lambda_1 = -3$$165\lambda_{1} = -495 \implies \lambda_{1} = -3$$ lambda_2 = 12(-3) + 38 = 2$$\lambda_{2} = 12(-3) + 38 = 2$$ Coordinates of C$C$: (11, 0, -5)$(11, 0, -5)$ Coordinates of D$D$: (8, 5, 11)$(8, 5, 11)$ ### Step 4: Calculate Magnitude squared left|overrightarrowCDright|^2 = (8 - 11)^2 + (5 - 0)^2 + (11 - (-5))^2$$\left|\overrightarrow{CD}\right|^{2} = (8 - 11)^2 + (5 - 0)^2 + (11 - (-5))^2$$ = (-3)^2 + 5^2 + 16^2$$= (-3)^2 + 5^2 + 16^2$$ = 9 + 25 + 256 = 290$$= 9 + 25 + 256 = 290$$ ### Pattern Recognition For intersecting lines via a transversal of known direction, represent intersection points generally using independent parameters lambda$\lambda$ and mu$\mu$. The difference vector MUST be proportional to the given direction ratio. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra

### Related Formula textShortest distance between two skew lines vecr = veca_1 + lambdavecb_1 text and vecr = veca_2 + muvecb_2 text is d = left| frac(veca_2 - veca_1) cdot (vecb_1 times vecb_2)|vecb_1 times vecb_2| right|$$\text{Shortest distance between two skew lines } \vec{r} = \vec{a_1} + \lambda\vec{b_1} \text{ and } \vec{r} = \vec{a_2} + \mu\vec{b_2} \text{ is } d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$$ ### Core Logic Let any point on the first line fracx - 12 = fracy + 1-3 = fracz1 = lambda$\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z}{1} = \lambda$ be given by P(2lambda + 1, -3lambda - 1, lambda)$P(2\lambda + 1, -3\lambda - 1, \lambda)$. The distance of P$P$ from the point A(1, -1, 0)$A(1, -1, 0)$ is 4sqrt14$4\sqrt{14}$. (2lambda + 1 - 1)^2 + (-3lambda - 1 + 1)^2 + (lambda - 0)^2 = (4sqrt14)^2$$(2\lambda + 1 - 1)^2 + (-3\lambda - 1 + 1)^2 + (\lambda - 0)^2 = (4\sqrt{14})^2$$ 4lambda^2 + 9lambda^2 + lambda^2 = 16 times 14$$4\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \times 14$$ 14lambda^2 = 224 implies lambda^2 = 16 implies lambda = pm 4$$14\lambda^2 = 224 \implies \lambda^2 = 16 \implies \lambda = \pm 4$$ ### Step 1: Finding Point P For lambda = 4$\lambda = 4$, point is P_1(9, -13, 4)$P_1(9, -13, 4)$. Distance from origin: sqrt81 + 169 + 16 = sqrt266$\sqrt{81 + 169 + 16} = \sqrt{266}$ For lambda = -4$\lambda = -4$, point is P_2(-7, 11, -4)$P_2(-7, 11, -4)$. Distance from origin: sqrt49 + 121 + 16 = sqrt186$\sqrt{49 + 121 + 16} = \sqrt{186}$ Since P$P$ is nearer to the origin, we choose lambda = -4$\lambda = -4$. Therefore, P(alpha, beta, gamma) = (-7, 11, -4)$P(\alpha, \beta, \gamma) = (-7, 11, -4)$. ### Step 2: Shortest Distance Calculation We need the shortest distance between Line 1: fracx + 71 = fracy - 112 = fracz + 43$\frac{x + 7}{1} = \frac{y - 11}{2} = \frac{z + 4}{3}$ and Line 2: fracx + 52 = fracy - 101 = fracz - 31$\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}$. Here, veca_1 = -7hati + 11hatj - 4hatk$\vec{a_1} = -7\hat{i} + 11\hat{j} - 4\hat{k}$ and vecb_1 = hati + 2hatj + 3hatk$\vec{b_1} = \hat{i} + 2\hat{j} + 3\hat{k}$. veca_2 = -5hati + 10hatj + 3hatk$\vec{a_2} = -5\hat{i} + 10\hat{j} + 3\hat{k}$ and vecb_2 = 2hati + hatj + hatk$\vec{b_2} = 2\hat{i} + \hat{j} + \hat{k}$. (veca_2 - veca_1) = 2hati - hatj + 7hatk$$(\vec{a_2} - \vec{a_1}) = 2\hat{i} - \hat{j} + 7\hat{k}$$ Shortest distance d$d$ is given by the determinant form: d = fracleft| beginmatrix 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 endmatrix right||vecb_1 times vecb_2|$$d = \frac{\left| \begin{matrix} 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} \right|}{|\vec{b_1} \times \vec{b_2}|}$$ Evaluate the determinant: = 2(2 - 3) - (-1)(1 - 6) + 7(1 - 4) = 2(-1) + 1(-5) + 7(-3) = -2 - 5 - 21 = -28$$= 2(2 - 3) - (-1)(1 - 6) + 7(1 - 4) = 2(-1) + 1(-5) + 7(-3) = -2 - 5 - 21 = -28$$ Now find |vecb_1 times vecb_2| = left| beginmatrix hati & hatj & hatk \\ 1 & 2 & 3 \\ 2 & 1 & 1 endmatrix right| = hati(2-3) - hatj(1-6) + hatk(1-4) = -hati + 5hatj - 3hatk$|\vec{b_1} \times \vec{b_2}| = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{matrix} \right| = \hat{i}(2-3) - \hat{j}(1-6) + \hat{k}(1-4) = -\hat{i} + 5\hat{j} - 3\hat{k}$ Magnitude is sqrt(-1)^2 + 5^2 + (-3)^2 = sqrt1 + 25 + 9 = sqrt35$\sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$. d = frac|-28|sqrt35 = frac28sqrt35 = frac4 times 7sqrt5 times 7 = frac4sqrt7sqrt5 = 4sqrtfrac75$$d = \frac{|-28|}{\sqrt{35}} = \frac{28}{\sqrt{35}} = \frac{4 \times 7}{\sqrt{5 \times 7}} = \frac{4\sqrt{7}}{\sqrt{5}} = 4\sqrt{\frac{7}{5}}$$ ### Pattern Recognition To find points on a line at a given distance from a fixed point on the line itself, the algebraic distance parameter lambda$\lambda$ translates to d^2 = lambda^2(a^2+b^2+c^2)$d^2 = \lambda^2(a^2+b^2+c^2)$, meaning 14lambda^2$14\lambda^2$ equates instantly to the squared given distance. Bypasses the complex distance formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textMidpoint M = left(fracx_1 + x_22, fracy_1 + y_22, fracz_1 + z_22right) text lies on the given line.$$\text{Midpoint } M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \text{ lies on the given line.}$$ textDirection ratio of PQ text is perpendicular to line's direction vector vecb, text so overrightarrowPQ cdot vecb = 0.$$\text{Direction ratio of } PQ \text{ is perpendicular to line's direction vector } \vec{b}, \text{ so } \overrightarrow{PQ} \cdot \vec{b} = 0.$$ ### Core Logic Given line L$L$: fracx - 63 = fracy - 72 = fracz - 7-2$\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$. Notice the standard form requires (z-7)/(-2)$(z-7)/(-2)$. Point P = (1, 2, a)$P = (1, 2, a)$ and its image is Q = (5, b, c)$Q = (5, b, c)$. The midpoint M$M$ of PQ$PQ$ must lie exactly on the line L$L$. M = left(frac1 + 52, frac2 + b2, fraca + c2right) = left(3, fracb + 22, fracc + a2right)$$M = \left(\frac{1 + 5}{2}, \frac{2 + b}{2}, \frac{a + c}{2}\right) = \left(3, \frac{b + 2}{2}, \frac{c + a}{2}\right)$$ ### Step 1: Using the Midpoint on the Line Substitute M$M$ into the line equation: frac3 - 63 = fracfracb + 22 - 72 = fracfracc + a2 - 7-2$$\frac{3 - 6}{3} = \frac{\frac{b + 2}{2} - 7}{2} = \frac{\frac{c + a}{2} - 7}{-2}$$ -1 = fracb - 124 = fracc + a - 14-4$$-1 = \frac{b - 12}{4} = \frac{c + a - 14}{-4}$$ From -1 = fracb - 124$-1 = \frac{b - 12}{4}$, we get -4 = b - 12 implies b = 8$-4 = b - 12 \implies b = 8$. From -1 = fracc + a - 14-4$-1 = \frac{c + a - 14}{-4}$, we get 4 = c + a - 14 implies c + a = 18$4 = c + a - 14 \implies c + a = 18$. ### Step 2: Using the Orthogonality Condition The vector overrightarrowPQ$\overrightarrow{PQ}$ must be perpendicular to the line's direction vector vecv = 3hati + 2hatj - 2hatk$\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$. overrightarrowPQ = (5 - 1)hati + (b - 2)hatj + (c - a)hatk = 4hati + 6hatj + (c - a)hatk$$\overrightarrow{PQ} = (5 - 1)\hat{i} + (b - 2)\hat{j} + (c - a)\hat{k} = 4\hat{i} + 6\hat{j} + (c - a)\hat{k}$$ (since b = 8$b = 8$) Now set the dot product to zero: 4(3) + 6(2) + (c - a)(-2) = 0$$4(3) + 6(2) + (c - a)(-2) = 0$$ 12 + 12 - 2(c - a) = 0 implies 24 = 2(c - a) implies c - a = 12$$12 + 12 - 2(c - a) = 0 \implies 24 = 2(c - a) \implies c - a = 12$$ ### Step 3: Solving for variables We have a system of linear equations: 1) c + a = 18$c + a = 18$ 2) c - a = 12$c - a = 12$ Adding both: 2c = 30 implies c = 15$2c = 30 \implies c = 15$. Substituting c$c$: 15 + a = 18 implies a = 3$15 + a = 18 \implies a = 3$. Therefore, a = 3, b = 8, c = 15$a = 3, b = 8, c = 15$. Calculate a^2 + b^2 + c^2$a^2 + b^2 + c^2$: a^2 + b^2 + c^2 = 3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$$a^2 + b^2 + c^2 = 3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$$

### Pattern Recognition Any 'image of a point' problem revolves around two strict constraints: 1) The line bisects the segment joining the point and its image (Midpoint lies on the line), and 2) The segment is orthogonal to the line (Dot product = 0). Directly imposing these generates decoupled simple linear equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry