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(D) The lines are $L_1: \frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $L_2: \frac{x-\lambda}{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\

Vedclass · Accepted Answer

$384$

Vedclass · Answer

(D) The lines are $L_1: \frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $L_2: \frac{x-\lambda}{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\sqrt{6}}{5}$.Points on the lines are $P_1(-\sqrt{6}, \sqrt{6}, \sqrt{6})$ and $P_2(\lambda, 2\sqrt{6}, -2\sqrt{6})$.Direction vectors are $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (3, 4, 5)$.The cross product is $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.The magnitude is $|\vec{n}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.The shortest distance is $d = \frac{|(\vec{P_2} - \vec{P_1}) \cdot \vec{n}|}{|\vec{n}|} = 6$.$\vec{P_2} - \vec{P_1} = (\lambda + \sqrt{6}, \sqrt{6}, -3\sqrt{6})$.$|(\lambda + \sqrt{6})(-1) + (\sqrt{6})(2) + (-3\sqrt{6})(-1)| = 6 	imes \sqrt{6} = 6\sqrt{6}$.$|-\lambda - \sqrt{6} + 2\sqrt{6} + 3\sqrt{6}| = 6\sqrt{6} \Rightarrow |4\sqrt{6} - \lambda| = 6\sqrt{6}$.Case $1: 4\sqrt{6} - \lambda = 6\sqrt{6} \Rightarrow \lambda = -2\sqrt{6}$.Case $2: 4\sqrt{6} - \lambda = -6\sqrt{6} \Rightarrow \lambda = 10\sqrt{6}$.Sum of values $= -2\sqrt{6} + 10\sqrt{6} = 8\sqrt{6}$.Square of the sum $= (8\sqrt{6})^2 = 64 	imes 6 = 384$.

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

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