A line passes through A(4, -6, -2)$A(4, -6, -2)$ and B(16, -2, 4)$B(16, -2, 4)$. The point P(a, b, c)$P(a, b, c)$ where a, b, c$a, b, c$ are non-negative integers, on the line AB$AB$ lies at a distance of 21$21$ units, from the point A$A$. The distance between the points P(a, b, c)$P(a, b, c)$ and Q(4, -12, 3)$Q(4, -12, 3)$ is equal to

### Related Formula Shortest distance between lines passing through veca_1, veca_2$\vec{a}_1, \vec{a}_2$ with directions vecp, vecq$\vec{p}, \vec{q}$: d = frac|(veca_2 - veca_1) cdot (vecp times vecq)||vecp times vecq|$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$$ ### Core Logic Identify parameters: A = (3, alpha, 3)$A = (3, \alpha, 3)$ and B = (-3, -7, beta) implies overrightarrowBA = 6hati + (alpha + 7)hatj + (3 - beta)hatk$B = (-3, -7, \beta) \implies \overrightarrow{BA} = 6\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k}$. Directions: vecp = 3hati - hatj + hatk$\vec{p} = 3\hat{i} - \hat{j} + \hat{k}$ and \ \vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}. Compute cross product $. Compute cross product $\vec{p} \times \vec{q}: $: $vecp times vecq = beginvmatrix hati & hatj & hatk \\ 3 & -1 & 1 \\ -3 & 2 & 4 endvmatrix = -6hati - 15hatj + 3hatk$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = -6hat{i} - 15hat{j} + 3hat{k}$ Magnitude $ Magnitude $|\vec{p} \times \vec{q}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}. ### Step 1: Apply Distance Equation Set shortest distance equation equal to $. ### Step 1: Apply Distance Equation Set shortest distance equation equal to $3\sqrt{30}: $: $frac|overrightarrowBA cdot (vecp times vecq)|3sqrt30 = 3sqrt30 implies |overrightarrowBA cdot (vecp times vecq)| = 270$\frac{|\overrightarrow{BA} \cdot (\vec{p} \times \vec{q})|}{3\sqrt{30}} = 3\sqrt{30} \implies |\overrightarrow{BA} \cdot (\vec{p} \times \vec{q})| = 270$ $ $-6(6) - 15(alpha + 7) + 3(3 - beta) = pm 270$-6(6) - 15(\alpha + 7) + 3(3 - \beta) = \pm 270$ $ $-36 - 15alpha - 105 + 9 - 3beta = pm 270 implies -132 - 15alpha - 3beta = pm 270$-36 - 15\alpha - 105 + 9 - 3\beta = \pm 270 \implies -132 - 15\alpha - 3\beta = \pm 270$ Choosing the negative branch for positive value extraction: $ Choosing the negative branch for positive value extraction: $-15alpha - 3beta = -138 implies 15alpha + 3beta = 138 implies 5alpha + beta = 46$-15\alpha - 3\beta = -138 \implies 15\alpha + 3\beta = 138 \implies 5\alpha + \beta = 46$$ ### Pattern Recognition Notice that the determinant logic perfectly structures linear equations. Simplifying the dot product using standard scaling helps prevent sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula Dot product of vectors: overrightarrowOA cdot overrightarrowOB = x_A x_B + y_A y_B + z_A z_B$$\overrightarrow{OA} \cdot \overrightarrow{OB} = x_A x_B + y_A y_B + z_A z_B$$ ### Core Logic Let any general point Q$Q$ on line L$L$ be (3lambda + 6, 2lambda + 7, -2lambda + 7)$(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$. Vector from P(1,2,3)$P(1,2,3)$ to Q$Q$ is overrightarrowPQ = (3lambda + 5)hati + (2lambda + 5)hatj + (-2lambda + 4)hatk$\overrightarrow{PQ} = (3\lambda + 5)\hat{i} + (2\lambda + 5)\hat{j} + (-2\lambda + 4)\hat{k}$. Since Q$Q$ is the foot of perpendicular, overrightarrowPQ cdot vecb = 0$\overrightarrow{PQ} \cdot \vec{b} = 0$, where vecb = 3hati + 2hatj - 2hatk$\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$: 3(3lambda + 5) + 2(2lambda + 5) - 2(-2lambda + 4) = 0$$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$$ 9lambda + 15 + 4lambda + 10 + 4lambda - 8 = 0 implies 17lambda = -17 implies lambda = -1$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \implies 17\lambda = -17 \implies \lambda = -1$$ Thus, the foot of perpendicular is Q(3, 5, 9)$Q(3, 5, 9)$. ### Step 1: Locate Points A and B Points A$A$ and B$B$ are on line L$L$, expressed via a distance parameter mu$\mu$ away from Q$Q$: General point expression from Q$Q$: A, B = (3mu + 3, 2mu + 5, -2mu + 9)$A, B = (3\mu + 3, 2\mu + 5, -2\mu + 9)$ [relative tracking points]. Distance squared = 68$= 68$: (3mu)^2 + (2mu)^2 + (-2mu)^2 = 68 implies 17mu^2 = 68 implies mu = pm 2$$(3\mu)^2 + (2\mu)^2 + (-2\mu)^2 = 68 \implies 17\mu^2 = 68 \implies \mu = \pm 2$$ For mu = 2$\mu = 2$: A = (3(2)+3, 2(2)+5, -2(2)+9) = (9, 9, 5)$A = (3(2)+3, 2(2)+5, -2(2)+9) = (9, 9, 5)$ For mu = -2$\mu = -2$: B = (3(-2)+3, 2(-2)+5, -2(-2)+9) = (-3, 1, 13)$B = (3(-2)+3, 2(-2)+5, -2(-2)+9) = (-3, 1, 13)$

### Step 2: Vector Dot Product Evaluation $overrightarrowOA cdot overrightarrowOB = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47$$$\overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47$$ ### Pattern Recognition When symmetrical points on a line are equidistant from a central foot position, utilizing standard parametric tracking simplifies vector resolution instantly. Double check calculations sequentially via parallel distance components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The direction vector of a line perpendicular to two given lines with direction vectors vecv_1$\vec{v}_1$ and \vec{v}_2 is determined by their cross product: vecv = vecv_1 times vecv_2$$\vec{v} = \vec{v}_1 \times \vec{v}_2$$ ### Core Logic The given point left(0, -frac12, 0right)$\left(0, -\frac{1}{2}, 0\right)$ lies on the required line: fracx - 1-2 = fracy + 4d = fracz - c-4$$\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$$ Substituting the point coordinates into the equation: frac0 - 1-2 = frac-frac12 + 4d = frac0 - c-4$$\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4}$$ frac12 = frac72d = fracc4 implies d = 7, quad c = 2$$\frac{1}{2} = \frac{7}{2d} = \frac{c}{4} \implies d = 7, \quad c = 2$$ ### Step 1: Cross Product Direction Ratios The direction vectors of the lines are vecv_1 = (1, a, b)$\vec{v}_1 = (1, a, b)$ and vecv_2 = (-b, a, 5)$\vec{v}_2 = (-b, a, 5)$. vecv = beginvmatrix hati & hatj & hatk \\ 1 & a & b \\ -b & a & 5 endvmatrix = hati(5a - ab) - hatj(5 + b^2) + hatk(a + ab)$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix} = \hat{i}(5a - ab) - \hat{j}(5 + b^2) + \hat{k}(a + ab)$$ Thus, the direction ratios of the line are proportional to: frac5a - ab-2 = frac-(b^2 + 5)7 = fraca + ab-4 quad dots text(i)$$\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{7} = \frac{a + ab}{-4} \quad \dots \text{(i)}$$ ### Step 2: Solve for a and b From the first and third components of equation (i): frac5a - ab-2 = fraca + ab-4 implies 2(5a - ab) = a + ab$$\frac{5a - ab}{-2} = \frac{a + ab}{-4} \implies 2(5a - ab) = a + ab$$ 10a - 2ab = a + ab implies 9a = 3ab implies b = 3$$10a - 2ab = a + ab \implies 9a = 3ab \implies b = 3$$ Now use the second component ratio with b = 3$b = 3$ and d = 7$d = 7$: frac-(3^2 + 5)7 = fraca + a(3)-4 implies frac-147 = frac4a-4 implies -2 = -a implies a = 2$$\frac{-(3^2 + 5)}{7} = \frac{a + a(3)}{-4} \implies \frac{-14}{7} = \frac{4a}{-4} \implies -2 = -a \implies a = 2$$ ### Step 3: Sum the Variables Summing up a, b, c, d$a, b, c, d$: a + b + c + d = 2 + 3 + 2 + 7 = 14$$a + b + c + d = 2 + 3 + 2 + 7 = 14$$ ### Pattern Recognition Substituting known point values into symmetric equations immediately determines structural values like c$c$ and d$d$ before running cross product systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The vector area of a triangle given two adjacent position vectors vecu$\vec{u}$ and vecv$\vec{v}$ is calculated as: textArea = frac12 |vecu times vecv|$$\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$$ ### Core Logic For line L_1$L_1$: fracx-11 = fracy-21 = fracz-01$\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let a general point be Q(lambda+1, lambda+2, lambda)$Q(\lambda+1, \lambda+2, \lambda)$. vecPQ = (lambda-4, lambda+1, lambda+3)$$\vec{PQ} = (\lambda-4, \lambda+1, \lambda+3)$$ Since vecPQ cdot vecm_1 = 0$\vec{PQ} \cdot \vec{m}_1 = 0$ (direction vector of L_1$L_1$ is (1,1,1)$(1,1,1)$): (lambda-4)(1) + (lambda+1)(1) + (lambda+3)(1) = 0 implies 3lambda = 0 implies lambda = 0$$(\lambda-4)(1) + (\lambda+1)(1) + (\lambda+3)(1) = 0 \implies 3\lambda = 0 \implies \lambda = 0$$ Thus, Q(1, 2, 0)$Q(1, 2, 0)$ and vecPQ = (-4, 1, 3)$\vec{PQ} = (-4, 1, 3)$.

### Step 1: Compute Foot R For line L_2$L_2$: fracx-21 = fracy1 = fracz-11$\frac{x-2}{1} = \frac{y}{1} = \frac{z-1}{1}$. Let a general point be R(mu+2, mu, mu+1)$R(\mu+2, \mu, \mu+1)$. vecPR = (mu-3, mu-1, mu+4)$$\vec{PR} = (\mu-3, \mu-1, \mu+4)$$ Since vecPR cdot vecm_2 = 0$\vec{PR} \cdot \vec{m}_2 = 0$ (direction vector of L_2$L_2$ is (1,1,1)$(1,1,1)$): (mu-3)(1) + (mu-1)(1) + (mu+4)(1) = 0 implies 3mu = 0 implies mu = 0$$(\mu-3)(1) + (\mu-1)(1) + (\mu+4)(1) = 0 \implies 3\mu = 0 \implies \mu = 0$$ Thus, R(2, 0, 1)$R(2, 0, 1)$ and vecPR = (-3, 1, 4)$\vec{PR} = (-3, 1, 4)$. ### Step 2: Area Vector Calculation The area A$A$ of Delta PQR$\Delta PQR$ is given by: A = frac12 |vecPQ times vecPR|$$A = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$$ vecPQ times vecPR = beginvmatrix hati & hatj & hatk \\ -4 & 1 & 3 \\ -3 & 1 & 4 endvmatrix = hati(4-3) - hatj(-16+9) + hatk(-4+3) = hati + 7hatj - hatk$$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 1 & 3 \\ -3 & 1 & 4 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-16+9) + \hat{k}(-4+3) = \hat{i} + 7\hat{j} - \hat{k}$$ textMagnitude squared: |vecPQ times vecPR|^2 = 1^2 + 7^2 + (-1)^2 = 1 + 49 + 1 = 51$$\text{Magnitude squared: } |\vec{PQ} \times \vec{PR}|^2 = 1^2 + 7^2 + (-1)^2 = 1 + 49 + 1 = 51$$ Let's re-verify the matrix arithmetic layout: vecPQ = (-4, 1, 3), vecPR = (-3, 1, 4)$$\vec{PQ} = (-4, 1, 3), \vec{PR} = (-3, 1, 4)$$ = 7hati + 7hatj + 7hatk$$= 7\hat{i} + 7\hat{j} + 7\hat{k}$$ |7(hati + hatj + hatk)|^2 = 49 cdot 3 = 147$$|7(\hat{i} + \hat{j} + \hat{k})|^2 = 49 \cdot 3 = 147$$ ### Step 3: Evaluate 4A^2 Since A = frac12 sqrt147$A = \frac{1}{2} \sqrt{147}$: 4A^2 = 4 cdot left(frac14 cdot 147right) = 147$$4A^2 = 4 \cdot \left(\frac{1}{4} \cdot 147\right) = 147$$ ### Pattern Recognition Setting up dot products systematically with general parametric forms quickly locks in spatial feet indices without complex geometric drawings. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula 1. Area of a \triangle with perpendicular height h$h$ and base b$b$: textArea = frac12 times b times h$$\text{Area} = \frac{1}{2} \times b \times h$$ 2. Since P$P$ is the reflection image of Q$Q$ across line L$L$, the line acts as a perpendicular bisector. For any point R$R$ lying on the line, the height from R$R$ to the line is RT$RT$, and the base QP = 2QT$QP = 2QT$. ### Core Logic Determine parameters for point R$R$ lying directly on line L [cite: 3405, 4071]: frac5-12 = fracp+13 = fracq4 Rightarrow 2 = fracp+13 = fracq4$$\frac{5-1}{2} = \frac{p+1}{3} = \frac{q}{4} \Rightarrow 2 = \frac{p+1}{3} = \frac{q}{4}$$ p+1 = 6 Rightarrow p = 5, quad q = 8 Rightarrow R = (5, 5, 8)$$p+1 = 6 \Rightarrow p = 5, \quad q = 8 \Rightarrow R = (5, 5, 8)$$ [cite: 4071, 4073]

### Step 1: Locate Foot of Perpendicular (T$T$) Let the foot of the perpendicular from Q(7, -2, 5)$Q(7, -2, 5)$ on line L$L$ be T(2lambda+1, 3lambda-1, 4lambda)$T(2\lambda+1, 3\lambda-1, 4\lambda)$ . The directional direction of L$L$ is vecb = 2hati + 3hatj + 4hatk$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ . Vector overrightarrowQT = (2lambda - 6)hati + (3lambda + 1)hatj + (4lambda - 5)hatk$\overrightarrow{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k}$ . Apply orthogonality condition overrightarrowQT cdot vecb = 0$\overrightarrow{QT} \cdot \vec{b} = 0$ : 2(2lambda - 6) + 3(3lambda + 1) + 4(4lambda - 5) = 0$$2(2\lambda - 6) + 3(3\lambda + 1) + 4(4\lambda - 5) = 0$$ 4lambda - 12 + 9lambda + 3 + 16lambda - 20 = 0 Rightarrow 29lambda - 29 = 0 Rightarrow lambda = 1$$4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \Rightarrow 29\lambda - 29 = 0 \Rightarrow \lambda = 1$$ [cite: 4077, 4078] Thus, T = (3, 2, 4)$T = (3, 2, 4)$. ### Step 2: Measure Geometric Distances Compute length QT$QT$ using distance metrics : QT = sqrt(3-7)^2 + (2 - (-2))^2 + (4-5)^2 = sqrt16 + 16 + 1 = sqrt33$$QT = \sqrt{(3-7)^2 + (2 - (-2))^2 + (4-5)^2} = \sqrt{16 + 16 + 1} = \sqrt{33}$$ Since P$P$ is the symmetrical image, base QP = 2QT = 2sqrt33$QP = 2QT = 2\sqrt{33}$. Compute length RT$RT$ representing height from vertex R(5, 5, 8)$R(5, 5, 8)$ to base line at T(3, 2, 4)$T(3, 2, 4)$ : RT = sqrt(5-3)^2 + (5-2)^2 + (8-4)^2 = sqrt4 + 9 + 16 = sqrt29$$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$$ ### Step 3: Calculate Squared Area Compute the \triangle area squared value [cite: 3406, 4081]: textArea = frac12 times QP times RT = frac12 times left(2sqrt33right) times sqrt29 = sqrt957$$\text{Area} = \frac{1}{2} \times QP \times RT = \frac{1}{2} \times \left(2\sqrt{33}\right) \times \sqrt{29} = \sqrt{957}$$ left(textArearight)^2 = 957$$\left(\text{Area}\right)^2 = 957$$ [cite: 4081, 4083] ### Pattern Recognition Because the image geometry creates an isosceles pairing from any point on the mirror line to the object and image point, the area reduces beautifully to 2 times textArea(triangle QTR) = QT times RT$2 \times \text{Area}(\triangle QTR) = QT \times RT$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry