Answer
$x+2y+4z = \frac{29}{2}$.
Work Step by Step
Let $\overrightarrow{PQ}=\langle 3,1,5\rangle - \langle 2,-1,1\rangle = \langle 1,2,4\rangle$ be a normal vector to the plane passing through points $P(2,-1,1)$ and $Q(3,1,5)$.
Let $\vec{n}=\langle 1,2,4\rangle$ be the normal vector to the plane, and let $\vec{r}_0=\left(\frac{5}{2},0,3\right)$ be the midpoint of $P$ and $Q$.
Then, the equation of the plane can be written in point-normal form as:
\begin{align*}
\vec{n}\cdot(\vec{r}-\vec{r}_0) &= 0 \\
\langle 1,2,4\rangle \cdot \left(\langle x,y,z\rangle - \left\langle \frac{5}{2},0,3\right\rangle\right) &= 0 \\
\left(x-\frac{5}{2}\right)+2y+4\left(z-3\right) &= 0 \\
x+2y+4z &= \frac{29}{2}.
\end{align*}
Therefore, the equation of the plane whose points are equidistant from $P$ and $Q$ is $x+2y+4z = \frac{29}{2}$.
