Let the given points be $P(-2,-1,-3)$ and $(1,-3,3)$ and the line $PQ$ be perpendicular to the plane.

Therefore $ PQ=(x_2-x_1),(y_2-y_1),(z_2-z_1)=(1-(-2)),(-3-(-1)),(3-(-3))$

$\qquad=(3,-2,6)$

It is given that $Q$ is the foot of the perpendicular to $P$ on the plane.

This implies that point Q lies on the plane.

Equation of the plane is $ax+by+cz+d$,where $(a,b,c)$ is normal.

$(ie) 3x-2y+6z+d=0$

Since Q lies on the plane,let us substitute for $(x,y,z)$

$(ie) ( 3 \times 1)+(-2) \times (-3)+(6 \times 3)+ d=0$

$(ie) 3+6+18+d=0$

$=> d=-27$

Therefore equation of the plane is

$3x-2y+6z-27=0$