Let S be the set of all values of lambda,for | vedclass

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(C) The lines are $L_1: \frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $L_2: \frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$.The points on the l

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$306$

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(C) The lines are $L_1: \frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $L_2: \frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$.The points on the lines are $A(\lambda, 3, -6)$ and $B(-\lambda, 0, 6)$. The vector $\vec{AB} = (-2\lambda, -3, 12)$.The direction vectors are $\vec{v_1} = (0, 4, 1)$ and $\vec{v_2} = (3, -4, 0)$.The cross product $\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 4 & 1 \ 3 & -4 & 0 \end{vmatrix} = \hat{i}(4) - \hat{j}(-3) + \hat{k}(-12) = 4\hat{i} + 3\hat{j} - 12\hat{k}$.The magnitude $|\vec{v_1} 	imes \vec{v_2}| = \sqrt{4^2 + 3^2 + (-12)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{v_1} 	imes \vec{v_2})|}{|\vec{v_1} 	imes \vec{v_2}|} = 13$.$|(-2\lambda, -3, 12) \cdot (4, 3, -12)| = 13 	imes 13 = 169$.$|-8\lambda - 9 - 144| = 169 \implies |-8\lambda - 153| = 169$.$8\lambda + 153 = 169$ or $8\lambda + 153 = -169$.$8\lambda = 16 \implies \lambda_1 = 2$.$8\lambda = -322 \implies \lambda_2 = -\frac{322}{8} = -40.25$.The sum $\sum_{\lambda \in S} \lambda = 2 - \frac{322}{8} = \frac{16 - 322}{8} = -\frac{306}{8}$.Then $8\left|\sum_{\lambda \in S} \lambdaight| = 8 	imes \left|-\frac{306}{8}ight| = 306$.

Let $S$ be the set of all values of $\lambda$,for which the shortest distance between the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum_{\lambda \in S} \lambda\right|$ is equal to

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