JEE Main 2026 (Online) 28th January Evening ShiftQuestion 1
If the distance of the point $P(43,\alpha,\beta)$, $\beta<0$, from the line $\vec r=4\hat i-\hat k+\mu(2\hat i+3\hat k)$, $\mu\in\mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13\sqrt{10}$, then $\alpha^2+\beta^2$ is equal to $\_\_\_\_\_\_$ .
3D Geometry
Answer$170$
Solution
Point on line: $(4+2\mu,0,-1+3\mu)=(43+3s,\alpha-s,\beta)$, so $\alpha=s$ and distance $=|s|\sqrt{10}=13\sqrt{10}$, $s=\pm13$. $s=13$ gives $\mu=39$, $\beta=116>0$ (rejected); $s=-13$ gives $\mu=0$, $\beta=-1$. So $\alpha^2+\beta^2=169+1=170$.