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Mathematics · 2025

JEE Main · 4 April 2025, Shift 1 · Q14

Let the shortest distance between the lines (x-3)/3=(y-α)/-1=(z-3)/1 and (x+3)/-3=(y+7)/2=(z-β)/4 be 3 √ 30. Then the positive value of 5 α+β is

Let the shortest distance between the lines $\displaystyle \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\displaystyle \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $\displaystyle 3 \sqrt{30}$. Then the positive value of $\displaystyle 5 \alpha+\beta$ is
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.