Answer
$2x-3y-9z=0$
Work Step by Step
$\vec{PQ}=(3-6)\textbf{i}+(2-1)\textbf{j}+(0-1)\textbf{k}= -3\textbf{i}+\textbf{j}-\textbf{k}$
$\vec{PR}=-6\textbf{i}-\textbf{j}-\textbf{k}$
$\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\-3&1&-1\\-6&-1&-1\end{vmatrix}=-2\textbf{i}+3\textbf{j}+9\textbf{k}$
$\vec{PQ}\times\vec{PR}$ is the normal vector to the plane through P, Q and R.
Using the formula for an equation of a plane, we have
$-2(x-0)+3(y-0)+9(z-0)=0$
or $2x-3y-9z=0$
