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(B) The given lines are $L_1: \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $L_2: \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$.The points on the

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$8$

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(B) The given lines are $L_1: \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $L_2: \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$.The points on the lines are $A(4, -1, 0)$ and $B(\lambda, -1, 2)$. The direction vectors are $\vec{d_1} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{d_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$.The shortest distance $d$ is given by $d = \left|\frac{(\vec{b}-\vec{a}) \cdot (\vec{d_1} 	imes \vec{d_2})}{|\vec{d_1} 	imes \vec{d_2}|}ight|$.First,calculate $\vec{d_1} 	imes \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = 2\hat{i} - \hat{j} + 0\hat{k}$.The magnitude is $|\vec{d_1} 	imes \vec{d_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.Now,$\vec{b}-\vec{a} = (\lambda-4)\hat{i} + 0\hat{j} + 2\hat{k}$.The dot product is $(\vec{b}-\vec{a}) \cdot (\vec{d_1} 	imes \vec{d_2}) = (\lambda-4)(2) + 0(-1) + 2(0) = 2(\lambda-4)$.Given the distance is $\frac{6}{\sqrt{5}}$,we have $\frac{|2(\lambda-4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$.This simplifies to $|2(\lambda-4)| = 6$,or $|\lambda-4| = 3$.Thus,$\lambda-4 = 3$ or $\lambda-4 = -3$,which gives $\lambda = 7$ or $\lambda = 1$.The sum of all possible values of $\lambda$ is $7 + 1 = 8$.

If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$,then the sum of all possible values of $\lambda$ is :

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