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Mathematics · 2026

JEE Main · 24 January 2026, Shift 2 · Q13

The sum of all values of α, for which the shortest distance between the lines (x+1)/(α)=(y-2)/-1=(z-4)/(-α) and x/(α)=(y-1)/2=(z-1)/(2 α) is √ 2, is

The sum of all values of $\displaystyle \alpha$, for which the shortest distance between the lines $\displaystyle \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$ and $\displaystyle \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2 \alpha}$ is $\displaystyle \sqrt{2}$, is
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.