Answer
$x+y+3z=5$
Work Step by Step
Step 1. Establish two vectors on the plane: $\vec u=\vec {PQ}=\langle 3-1,-4-1,2-1 \rangle=\langle 2,-5,1 \rangle$ and $\vec v=\vec {PR}=\langle 6-1,-1-1,0-1 \rangle=\langle 5,-2,-1 \rangle$
Step 2. Find a normal to the plane with the cross product of the above two vectors: $\vec u\times\vec v=\langle (5+2),(5+2),(-4+25) \rangle=\langle 7,7,21 \rangle$. Simplify this vector as $\vec n=\langle 1,1,3 \rangle$
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: $(x-1)+(y-1)+3(z-1)=0$ or $x+y+3z=5$
