Let the lines l_1: frac{x+5}{3}=frac{y+4}{1}= | vedclass

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(C) The line $l_2$ is given by the intersection of two planes: $3x+2y+z-2=0$ and $x-3y+2z-13=0$. The direction vector of $l_2$ is $\vec{v} = \begin{vm

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$10$

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(C) The line $l_2$ is given by the intersection of two planes: $3x+2y+z-2=0$ and $x-3y+2z-13=0$. The direction vector of $l_2$ is $\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 1 \ 1 & -3 & 2 \end{vmatrix} = 7\hat{i} - 5\hat{j} - 11\hat{k}$.Since $l_1$ and $l_2$ are coplanar,the scalar triple product of the vector connecting points on the lines and the direction vectors must be zero. Solving for $\alpha$,we find $\alpha = 7$.Now,$l_1$ is $\frac{x+5}{3}=\frac{y+4}{1}=\frac{z-7}{-2} = \lambda$. Thus,any point $P$ on $l_1$ is $(3\lambda-5, \lambda-4, -2\lambda+7)$.The vector $\vec{PQ} = (3\lambda-5 - (-4), \lambda-4 - (-3), -2\lambda+7 - 2) = (3\lambda-1, \lambda-1, -2\lambda+5)$.Since $PQ \perp l_1$,the dot product of $\vec{PQ}$ and the direction vector of $l_1$ $(3, 1, -2)$ is zero:$3(3\lambda-1) + 1(\lambda-1) - 2(-2\lambda+5) = 0 \Rightarrow 9\lambda - 3 + \lambda - 1 + 4\lambda - 10 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.Substituting $\lambda=1$,we get $P(-2, -3, 5)$.Thus,$|a|+|b|+|c| = |-2| + |-3| + |5| = 2+3+5 = 10$.

Let the lines $l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$ and $l_2: 3x+2y+z-2=0=x-3y+2z-13$ be coplanar. If the point $P(a, b, c)$ on $l_1$ is nearest to the point $Q(-4, -3, 2)$,then $|a|+|b|+|c|$ is equal to

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