JEE Main 2022 (Online) 29th July Evening ShiftQuestion 1
If $(2,3,9)$, $(5,2,1)$, $(1,\lambda,8)$ and $(\lambda,2,3)$ are coplanar, then the product of all possible values of $\lambda$ is :
3D Geometry
Solution
Vectors from $(2,3,9)$: $(3,-1,-8)$, $(-1,\lambda-3,-1)$, $(\lambda-2,-1,-6)$. Coplanarity: determinant $=0$. Expanding the determinant gives $8\lambda^2-57\lambda+95=0$ up to a nonzero constant factor (roots $\frac{57\pm\sqrt{209}}{16}$, confirmed with sympy). Product of roots $=\frac{95}{8}$. Option (D).