Question

Find the equation of the plane passing through the points

Answer 1

Extract the normal vectors from the given planes:

$-2x+y+z+2=0\implies\vec n_1=(-2,1,1)$

$x+y-3z+1=0\implies\vec n_2=(1,1,-3)$

(which are unique up to their signs, meaning either $\vec n_1$ or $-\vec n_1$ are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both $\vec n_1$ and $\vec n_2$, which in turn means its own normal vector $\vec n_3$ should be perpendicular to both $\vec n_1$ and $\vec n_2$.

Enter the cross product:

$\vec n_3=\vec n_1\times\vec n_2=(-4,-5,-3)$

or (4, 5, 3), which also works.

The given plane passes through (-1, 1, 4), so its equation is

$(x+1,y-1,z-4)\cdot\vec n_3=0$

Simplify:

$(x+1,y-1,z-4)\cdot(4,5,3)=0$

$4(x+1)+5(y-1)+3(z-4)=0$

$\boxed{4x+5y+3z=13}$