Mathematics · 2024
JEE Main · 5 April 2024, Shift 1 · Q17
Let d be the distance of the point of intersection of the lines (x+6)/3=y/2=(z+1)/1 and (x-7)/4=(y-9)/3=(z-4)/2 from the point (7,8,9). Then d^2+6 is…
Let $\displaystyle d$ be the distance of the point of intersection of the lines $\displaystyle \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\displaystyle \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $\displaystyle (7,8,9)$. Then $\displaystyle \mathrm{d}^2+6$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 75$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.