JEE Advanced 2010 Paper 1Question 1
Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is
3D Geometry
Solution
Normal of the plane of the two lines: $(3,4,2)\times(4,2,3)=(8,-1,-10)$. Required normal is perpendicular to $(2,3,4)$ and $(8,-1,-10)$: $(-26,52,-26)\propto(1,-2,1)$, so the plane is $x-2y+z=0$.