Mathematics · 2025
JEE Main · 4 April 2025, Shift 2 · Q15
Let A be the point of intersection of the lines L_1: (x-7)/1=(y-5)/0=(z-3)/-1 and L_2: (x-1)/3=(y+3)/4=(z+7)/5. Let B and C be the points on the…
Let $\displaystyle A$ be the point of intersection of the lines $\displaystyle \mathrm{L}_1: \frac{x-7}{1}=\frac{y-5}{0}=\frac{z-3}{-1}$ and $\displaystyle \mathrm{L}_2: \frac{x-1}{3}=\frac{y+3}{4}=\frac{z+7}{5}$. Let B and C be the points on the lines $\displaystyle \mathrm{L}_1$ and $\displaystyle \mathrm{L}_2$ respectively such that $\displaystyle \mathrm{AB}=\mathrm{AC}=\sqrt{15}$. Then the square of the area of the triangle ABC is :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 54$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.