### Core Logic A vector in the direction of the required line is perpendicular to both given lines. We obtain it via cross product of their direction vectors: vecn = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 5 \\ -1 & 3 & 2 endvmatrix = -9hati - 9hatj + 9hatk$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k}$ Taking the direction vector as hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$. ### Step 1: Required Line Equation The line passes through P(5, -4, 3)$P(5, -4, 3)$ with direction hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$. Equation: vecr = (5hati - 4hatj + 3hatk) + alpha(hati + hatj - hatk)$\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \alpha(\hat{i} + \hat{j} - \hat{k})$. ### Step 2: Projection & Distance Any point on the line is M(5+alpha, -4+alpha, 3-alpha)$M(5+\alpha, -4+\alpha, 3-\alpha)$. We need distance from Q(0, 2, -2)$Q(0, 2, -2)$. Vector vecQM = (5+alpha)hati + (alpha-6)hatj + (5-alpha)hatk$\vec{QM} = (5+\alpha)\hat{i} + (\alpha-6)\hat{j} + (5-\alpha)\hat{k}$. Since vecQM$\vec{QM}$ is perpendicular to the line direction (hati + hatj - hatk)$(\hat{i} + \hat{j} - \hat{k})$: (5+alpha)(1) + (alpha-6)(1) + (5-alpha)(-1) = 0$(5+\alpha)(1) + (\alpha-6)(1) + (5-\alpha)(-1) = 0$ 5 + alpha + alpha - 6 - 5 + alpha = 0 implies 3alpha = 6 implies alpha = 2.$5 + \alpha + \alpha - 6 - 5 + \alpha = 0 \implies 3\alpha = 6 \implies \alpha = 2.$

### Step 3: Distance calculation Substitute alpha = 2$\alpha = 2$ in vecQM$\vec{QM}$: vecQM = 7hati - 4hatj + 3hatk$\vec{QM} = 7\hat{i} - 4\hat{j} + 3\hat{k}$. Distance |vecQM| = sqrt7^2 + (-4)^2 + 3^2 = sqrt49 + 16 + 9 = sqrt74$|\vec{QM}| = \sqrt{7^2 + (-4)^2 + 3^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra

### Core Logic Line 1: fracx1 = fracy1 = fracz-10 = r implies Q(r, r, 1)$\frac{x}{1} = \frac{y}{1} = \frac{z-1}{0} = r \implies Q(r, r, 1)$. Line 2: fracx1 = fracy-1 = fracz+10 = k implies R(k, -k, -1)$\frac{x}{1} = \frac{y}{-1} = \frac{z+1}{0} = k \implies R(k, -k, -1)$. ### Step 1: Perpendicular Conditions Vector vecPQ = (r-a)hati + (r-a)hatj + (1-a)hatk$\vec{PQ} = (r-a)\hat{i} + (r-a)\hat{j} + (1-a)\hat{k}$. vecPQ cdot textDirection of Line 1 = 0 implies (r-a)(1) + (r-a)(1) + (1-a)(0) = 0$\vec{PQ} \cdot \text{Direction of Line 1} = 0 \implies (r-a)(1) + (r-a)(1) + (1-a)(0) = 0$. 2r - 2a = 0 implies r = a$2r - 2a = 0 \implies r = a$. Thus, vecPQ = 0hati + 0hatj + (1-a)hatk$\vec{PQ} = 0\hat{i} + 0\hat{j} + (1-a)\hat{k}$. Vector vecPR = (k-a)hati + (-k-a)hatj + (-1-a)hatk$\vec{PR} = (k-a)\hat{i} + (-k-a)\hat{j} + (-1-a)\hat{k}$. vecPR cdot textDirection of Line 2 = 0 implies (k-a)(1) + (-k-a)(-1) + (-1-a)(0) = 0$\vec{PR} \cdot \text{Direction of Line 2} = 0 \implies (k-a)(1) + (-k-a)(-1) + (-1-a)(0) = 0$. k - a + k + a = 0 implies 2k = 0 implies k = 0$k - a + k + a = 0 \implies 2k = 0 \implies k = 0$. Thus, vecPR = -ahati - ahatj - (a+1)hatk$\vec{PR} = -a\hat{i} - a\hat{j} - (a+1)\hat{k}$. ### Step 2: Right Angle Condition Given angle QPR = 90^circ implies vecPQ cdot vecPR = 0$\angle QPR = 90^\circ \implies \vec{PQ} \cdot \vec{PR} = 0$. (0)(-a) + (0)(-a) + (1-a)(-(a+1)) = 0$(0)(-a) + (0)(-a) + (1-a)(-(a+1)) = 0$ -(1-a)(1+a) = 0 implies a^2 - 1 = 0 implies a^2 = 1$-(1-a)(1+a) = 0 \implies a^2 - 1 = 0 \implies a^2 = 1$ Therefore, 12a^2 = 12(1) = 12$12a^2 = 12(1) = 12$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra