Answer
$15x-20y-12z=60$
Work Step by Step
Step 1. Establish two vectors on the plane: $\vec u=\vec {PQ}=\langle 0-4,-3-0,0-0 \rangle=\langle -4,-3,0 \rangle$ and $\vec v=\vec {PR}=\langle 0-4,0-0,-5-0 \rangle=\langle -4,0,-5 \rangle$
Step 2. Find a normal to the plane with the cross product of the above two vectors: $\vec n=\vec u\times\vec v=\langle (15-0),(0-20),(0-12) \rangle=\langle 15,-20,-12\rangle$.
Step 3. The general equation for a plane containing point $P(x_0,y_0,z_0)$ and having a normal $\vec n=\langle a,b,c \rangle$ is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Step 4. The equation of the plane in the problem can be written as: $15(x-4)-20(y-0)-12(z-0)=0$ or $15x-20y-12z=60$
