If the image of the point (4, 4, 3)$(4, 4, 3)$ in the line fracx - 12 = fracy - 21 = fracz - 13$\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is (alpha, beta, gamma)$(\alpha, \beta, \gamma)$, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to (1) 9 (2) 12 (3) 8 (4) 7

### Related Formula For three points P$P$, A$A$, and B$B$ to be collinear, their direction vectors must be proportional: vecPA parallel vecPB implies fracx_A - x_Px_B - x_P = fracy_A - y_Py_B - y_P = fracz_A - z_Pz_B - z_P$$\vec{PA} \parallel \vec{PB} \implies \frac{x_A - x_P}{x_B - x_P} = \frac{y_A - y_P}{y_B - y_P} = \frac{z_A - z_P}{z_B - z_P}$$

### Core Logic Express general coordinates for A$A$ on L_1$L_1$ and B$B$ on L_2$L_2$ [cite: 1204]: L_1: fracx-12=fracy-23=fracz-34=p implies A(2p+1, 3p+2, 4p+3)$$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p \implies A(2p+1, 3p+2, 4p+3)$$ [cite: 1204, 1206] L_2: fracx-61=fracy1=fracz-4-1=q implies B(q+6, q, 4-q)$$L_2: \frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q \implies B(q+6, q, 4-q)$$ [cite: 1204, 1207] Direction ratios (D.R.) from P(4,1,0)$P(4,1,0)$ [cite: 1208, 1209]: textD.R. of PA = (2p-3, 3p+1, 4p+3)$$\text{D.R. of } PA = (2p-3, 3p+1, 4p+3)$$ [cite: 1208] textD.R. of PB = (q+2, q-1, 4-q)$$\text{D.R. of } PB = (q+2, q-1, 4-q)$$ [cite: 1209] Since P, A, B$P, A, B$ lie on the same line [cite: 1210]: frac2p-3q+2 = frac3p+1q-1 = frac4p+34-q$$\frac{2p-3}{q+2} = \frac{3p+1}{q-1} = \frac{4p+3}{4-q}$$ [cite: 1210] ### Step 1: Solving the system of equations Equating expressions to solve for parameters p$p$ and q$q$ [cite: 1211, 1212]: From first two [cite: 1211]: 2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 implies pq + 8p + 4q - 1 = 0$$2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 \implies pq + 8p + 4q - 1 = 0$$ [cite: 1211, 1212] From second and third components [cite: 1212]: 12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 implies 7pq - 16p + 4q - 7 = 0$$12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 \implies 7pq - 16p + 4q - 7 = 0$$ [cite: 1212] Solving equations simultaneously gives [cite: 1234]: p = -1, quad q = 3$$p = -1, \quad q = 3$$ [cite: 1234] Substituting parameters back yields the points [cite: 1236]: A(-1, -1, -1), quad B(9, 3, 1)$$A(-1, -1, -1), \quad B(9, 3, 1)$$ [cite: 1236] ### Step 2: Evaluating the Determinant Substitute values of A$A$ and B$B$ into the matrix grid [cite: 1244]: beginvmatrix 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 endvmatrix$$\begin{vmatrix} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{vmatrix}$$ [cite: 1244] Applying row/column operations (C_3 rightarrow C_3 - C_1$C_3 \rightarrow C_3 - C_1$) [cite: 1244]: beginvmatrix 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 endvmatrix = 1((-1)(-8) - 0) = 8$$\begin{vmatrix} 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 \end{vmatrix} = 1((-1)(-8) - 0) = 8$$ [cite: 1244] ### Pattern Recognition Collinearity problem involving parameter matches across skew line orientations. Reducing fractions systematically prevents higher-degree calculation mistakes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Determinants

### Related Formula Shortest distance between perpendicular axes vectors: textS.D. = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$$\text{S.D.} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$ ### Core Logic Represent equations of straight lines symmetrically using vector directions [cite: 1418]: L_1: fracx-10 = fracy-20 = fracz-31$$L_1: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$$ [cite: 1418] L_2: fracx-lambda0 = fracy-51 = fracz-60$$L_2: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$$ [cite: 1418] Evaluating the standard shortest distance configuration formula[cite: 1419, 1420]: textS.D. = |lambda - 1| = 3 implies lambda - 1 = pm 3$$\text{S.D.} = |\lambda - 1| = 3 \implies \lambda - 1 = \pm 3$$ [cite: 1420] lambda = 4 quad textor quad lambda = -2$$\lambda = 4 \quad \text{or} \quad \lambda = -2$$ [cite: 1420] Given the condition lambda_2 < lambda_1$\lambda_2 < \lambda_1$ [cite: 698]: lambda_1 = 4, quad lambda_2 = -2$$\lambda_1 = 4, \quad \lambda_2 = -2$$ [cite: 1421, 1422] ### Step 1: Distance calculation from line We need to find the square of distance from point P(4, -2, 7)$P(4, -2, 7)$ to line L_1$L_1$ [cite: 1424]. Any general matching point coordinates tracking along path L_1$L_1$ look like Q(1, 2, t+3)$Q(1, 2, t+3)$ [cite: 1424]. Form a perpendicular projection vector condition [cite: 1425]: vecPQ = (-3, 4, t-4)$$\vec{PQ} = (-3, 4, t-4)$$ [cite: 1425] Since vecPQ cdot hatk = 0 implies t-4 = 0 implies t=4$\vec{PQ} \cdot \hat{k} = 0 \implies t-4 = 0 \implies t=4$ [cite: 1425, 1426]. Thus, the foot of perpendicular is Q(1, 2, 7)$Q(1, 2, 7)$ [cite: 1427]. Evaluate the squared distance component magnitude [cite: 1427]: PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$$ [cite: 1427, 1428] ### Pattern Recognition For lines parallel directly to independent Cartesian coordinate grid lines, the shortest paths are simply direct plane projections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The shortest distance between two lines vecr = veca_1 + lambda vecb_1$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and vecr = veca_2 + mu vecb_2$\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by: d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$ ### Core Logic From the given line equations: Line 1 passes through veca_1 = -hati + 0hatj + 0hatk$\vec{a}_1 = -\hat{i} + 0\hat{j} + 0\hat{k}$ along vector vecb_1 = 3hati + 4hatj + 5hatk$\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$. Line 2 passes through veca_2 = phati + 2hatj + hatk$\vec{a}_2 = p\hat{i} + 2\hat{j} + \hat{k}$ along vector vecb_2 = 2hati + 3hatj + 4hatk$\vec{b}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Vector difference: veca_2 - veca_1 = (p + 1)hati + 2hatj + hatk$$\vec{a}_2 - \vec{a}_1 = (p + 1)\hat{i} + 2\hat{j} + \hat{k}$$ Computing the cross product vecb_1 times vecb_2$\vec{b}_1 \times \vec{b}_2$: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 3 & 4 & 5 \\ 2 & 3 & 4 endvmatrix = hati(16-15) - hatj(12-10) + hatk(9-8) = hati - 2hatj + hatk$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$$ Magnitude |vecb_1 times vecb_2| = sqrt1^2 + (-2)^2 + 1^2 = sqrt6$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$. ### Step 1: Applying the Shortest Distance Value Substitute these into the distance equation: d = frac|big((p + 1)hati + 2hatj + hatkbig) cdot big(hati - 2hatj + hatkbig)|sqrt6 = frac1sqrt6$$d = \frac{|\big((p + 1)\hat{i} + 2\hat{j} + \hat{k}\big) \cdot \big(\hat{i} - 2\hat{j} + \hat{k}\big)|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$ |(p + 1)(1) + 2(-2) + 1(1)| = 1$$|(p + 1)(1) + 2(-2) + 1(1)| = 1$$ |p + 1 - 4 + 1| = 1 implies |p - 2| = 1$$|p + 1 - 4 + 1| = 1 \implies |p - 2| = 1$$ This yields two values for p$p$: - p - 2 = 1 implies p = 3$p - 2 = 1 \implies p = 3$ - p - 2 = -1 implies p = 1$p - 2 = -1 \implies p = 1$ Given that a, b$a, b$ are the parameters with a < b$a < b$, we assign a = 1$a = 1$ and b = 3$b = 3$. ### Step 2: Computing Latus Rectum of the Ellipse The ellipse equation is: fracx^21^2 + fracy^23^2 = 1$$\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$$ Since b > a$b > a$, the formula for the length of the latus rectum is: textLatus Rectum = frac2a^2b = frac2(1)^23 = frac23$$\text{Latus Rectum} = \frac{2a^2}{b} = \frac{2(1)^2}{3} = \frac{2}{3}$$ ### Pattern Recognition Be careful with coordinate geometry variables; when an ellipse satisfies b > a$b > a$, the major axis is along the y-axis, making the latus rectum equal to frac2a^2b$\frac{2a^2}{b}$ instead of frac2b^2a$\frac{2b^2}{a}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Conic Sections

### Core Logic First, find the point of intersection A$A$ by solving the lines. Any point on L_1$L_1$ can be written as (lambda + 7, 5, -lambda + 3)$(\lambda + 7, 5, -\lambda + 3)$. Substituting this point into the equation for L_2$L_2$: frac(lambda + 7) - 13 = frac5 + 34 implies fraclambda + 63 = 2 implies lambda = 0$$\frac{(\lambda + 7) - 1}{3} = \frac{5 + 3}{4} \implies \frac{\lambda + 6}{3} = 2 \implies \lambda = 0$$ Thus, the intersection point is A = (7, 5, 3)$A = (7, 5, 3)$. ### Step 1: Calculating the Angle between lines The directional vectors of lines L_1$L_1$ and L_2$L_2$ are vecu = hati - hatk$\vec{u} = \hat{i} - \hat{k}$ and vecv = 3hati + 4hatj + 5hatk$\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively. costheta = frac|vecu cdot vecv||vecu||vecv| = frac|1(3) + 0(4) - 1(5)|sqrt1^2+(-1)^2 sqrt3^2+4^2+5^2 = frac|3 - 5|sqrt2sqrt50 = frac210 = frac15$$\cos\theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}||\vec{v}|} = \frac{|1(3) + 0(4) - 1(5)|}{\sqrt{1^2+(-1)^2} \sqrt{3^2+4^2+5^2}} = \frac{|3 - 5|}{\sqrt{2}\sqrt{50}} = \frac{2}{10} = \frac{1}{5}$$ Now, find sintheta$\sin\theta$: sintheta = sqrt1 - cos^2theta = sqrt1 - frac125 = fracsqrt245$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{1}{25}} = \frac{\sqrt{24}}{5}$$

### Step 2: Finding Area of the Triangle The area of triangle ABC$\triangle ABC$ given two sides and their included angle is: textArea = frac12 cdot AB cdot AC cdot sintheta$$\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin\theta$$ Given AB = AC = sqrt15$AB = AC = \sqrt{15}$: textArea = frac12 cdot sqrt15 cdot sqrt15 cdot fracsqrt245 = frac15sqrt2410 = frac3sqrt242$$\text{Area} = \frac{1}{2} \cdot \sqrt{15} \cdot \sqrt{15} \cdot \frac{\sqrt{24}}{5} = \frac{15\sqrt{24}}{10} = \frac{3\sqrt{24}}{2}$$ Squaring the area: textArea^2 = left(frac3sqrt242right)^2 = frac9 times 244 = 9 times 6 = 54$$\text{Area}^2 = \left(\frac{3\sqrt{24}}{2}\right)^2 = \frac{9 \times 24}{4} = 9 \times 6 = 54$$ ### Pattern Recognition Since B$B$ and C$C$ lie on lines intersecting at A$A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Properties of Triangles