Mathematics · 2026
JEE Main · 22 January 2026, Shift 2 · Q15
Let L be the line (x+1)/2=(y+1)/3=(z+3)/6 and let S be the set of all points ( a, b, c ) on L, whose distance from the line (x+1)/2=(y+1)/3=(z-9)/0…
Let L be the line $\displaystyle \frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $\displaystyle (\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L , whose distance from the line $\displaystyle \frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line L is 7. Then $\displaystyle \sum_{(\mathrm{a}, \mathrm{b}, \mathrm{c}) \in \mathrm{S}}(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 34$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.