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(B) Let the given line be $L_1: \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1} = \lambda$. Any point $M$ on $L_1$ is $(2\lambda-1, \lambda+3, -\lambda-2)$

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(B) Let the given line be $L_1: \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1} = \lambda$. Any point $M$ on $L_1$ is $(2\lambda-1, \lambda+3, -\lambda-2)$.Given point $P = (2,3,1)$. The vector $\vec{PM} = (2\lambda-3, \lambda, -\lambda-3)$.Since $\vec{PM}$ is perpendicular to the direction vector of $L_1$,which is $\vec{v_1} = (2, 1, -1)$:$2(2\lambda-3) + 1(\lambda) - 1(-\lambda-3) = 0$$4\lambda - 6 + \lambda + \lambda + 3 = 0 \Rightarrow 6\lambda = 3 \Rightarrow \lambda = \frac{1}{2}$.Thus,$M = (0, \frac{7}{2}, -\frac{5}{2})$.Let $P'(x', y', z')$ be the mirror image of $P$ with respect to $L_1$. Since $M$ is the midpoint of $PP'$:$\frac{x'+2}{2} = 0 \Rightarrow x' = -2$$\frac{y'+3}{2} = \frac{7}{2} \Rightarrow y' = 4$$\frac{z'+1}{2} = -\frac{5}{2} \Rightarrow z' = -6$So,$P' = (-2, 4, -6)$.The plane contains the line $L_2: \frac{x-2}{3} = \frac{y-1}{-2} = \frac{z+1}{1}$.The plane passes through $P'(-2, 4, -6)$ and a point $A(2, 1, -1)$ on $L_2$,and is parallel to the direction vector of $L_2$,$\vec{v_2} = (3, -2, 1)$.The vector $\vec{P'A} = (2 - (-2), 1 - 4, -1 - (-6)) = (4, -3, 5)$.The normal to the plane is $\vec{n} = \vec{P'A} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -3 & 5 \ 3 & -2 & 1 \end{vmatrix} = \hat{i}(-3+10) - \hat{j}(4-15) + \hat{k}(-8+9) = 7\hat{i} + 11\hat{j} + 1\hat{k}$.The equation of the plane is $7(x-2) + 11(y-1) + 1(z+1) = 0 \Rightarrow 7x + 11y + z = 14 + 11 - 1 = 24$.Comparing with $\alpha x + \beta y + \gamma z = 24$,we get $\alpha=7, \beta=11, \gamma=1$.Therefore,$\alpha+\beta+\gamma = 7+11+1 = 19$.

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