JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
Let a line passing through the point $(4,1,0)$ intersect the line $L_1:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha,\beta,\gamma)$ and the line $L_2:x-6=y=-z+4$ at the point $B(a,b,c)$. Then $\begin{vmatrix}1&0&1\\\alpha&\beta&\gamma\\a&b&c\end{vmatrix}$ is equal to
3D Geometry
Solution
$A=(1+2s,2+3s,3+4s)$, $B=(6+u,u,4-u)$, with $P=(4,1,0)$, $A,B,P$ collinear: $\vec{PA}\times\vec{PB}=0$ gives $s=-1,\ u=3$. $A=(-1,-1,-1)$, $B=(9,3,1)$. $\begin{vmatrix}1&0&1\\-1&-1&-1\\9&3&1\end{vmatrix}=1(-1+3)-0+1(-3+9)=2+6=8$. Option (A).