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The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\sqrt{2}$, is
[JEE MAIN 24 Jan 2026 S II]
(1) 8
(2) -6
(3) 6
(4) -8Solution
First line $\mathrm{L}_{1}$ passes through point $\mathrm{A}(-1,2,4)$ with drs $\alpha,-1,-\alpha$
First line $\mathrm{L}_{1}$ passes through point $\mathrm{B}(0,1,1)$ with drs $\alpha, 2,2 \alpha$
shortest distance $\mathrm{b} / \mathrm{w}$ skew lines $=\frac{\left|\begin{array}{ccc}\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right|}{\left\|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right\|}$
$=\frac{\left|\begin{array}{ccc}1 & -1 & -3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}{\left\|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right\|}=\sqrt{2}$
$\Rightarrow \frac{\left|(-2 \alpha+2 \alpha)+1\left(2 \alpha^{2}+\alpha^{2}\right)-3(3 \alpha)\right|}{\left|0-3 \alpha^{2} \hat{\mathrm{j}}+3 \alpha \hat{\mathrm{k}}\right|}=\sqrt{2}$
$\Rightarrow \frac{\left|3 \alpha^{2}-9 \alpha\right|}{\sqrt{9 \alpha^{4}+9 \alpha^{2}}}=\sqrt{2} \Rightarrow \frac{3|\alpha-3||\alpha|}{3 \alpha \sqrt{\alpha^{2}+1}}=\sqrt{2}$
$\Rightarrow|\alpha-3|=\sqrt{2} \sqrt{\alpha^{2}+1} \Rightarrow \alpha^{2}+9-6 \alpha=2 \alpha^{2}+2$
$\Rightarrow \alpha^{2}+6 \alpha-7=0 \Rightarrow(\alpha+7)(\alpha-1)=0$
$\alpha=-7,1$ sum of $\alpha=-6$ Answer(2)
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