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(D) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$.Points on the lines ar

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$-3$

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(D) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$.Points on the lines are $A(1, 2, 3)$ and $B(0, 0, 5)$.Direction vectors are $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b_2} = \hat{i} + \alpha\hat{j} + \hat{k}$.The cross product $\vec{n} = \vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 1 & \alpha & 1 \end{vmatrix} = \hat{i}(3-4\alpha) - \hat{j}(2-4) + \hat{k}(2\alpha-3) = (3-4\alpha)\hat{i} + 2\hat{j} + (2\alpha-3)\hat{k}$.Vector $\vec{AB} = (0-1)\hat{i} + (0-2)\hat{j} + (5-3)\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k}$.Shortest distance $d = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|} = \frac{|(-1)(3-4\alpha) + (-2)(2) + (2)(2\alpha-3)|}{\sqrt{(3-4\alpha)^2 + 2^2 + (2\alpha-3)^2}} = \frac{5}{\sqrt{6}}$.$|4\alpha - 3 - 4 + 4\alpha - 6| = |8\alpha - 13|$.$\frac{|8\alpha - 13|}{\sqrt{16\alpha^2 - 24\alpha + 9 + 4 + 4\alpha^2 - 12\alpha + 9}} = \frac{5}{\sqrt{6}}$.$\frac{|8\alpha - 13|}{\sqrt{20\alpha^2 - 36\alpha + 22}} = \frac{5}{\sqrt{6}}$.Squaring both sides: $6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22)$.$384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550$.$116\alpha^2 + 348\alpha - 464 = 0$.Dividing by $116$: $\alpha^2 + 3\alpha - 4 = 0$.Sum of roots $\alpha_1 + \alpha_2 = -3$.

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

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