Answer
$$ y=1.$$
Work Step by Step
We have
$$\overrightarrow{PQ}=Q-P=(1,1,2)-(5,1,1)=\langle -4,0,1\rangle,$$
$$\overrightarrow{PR}=R-P=(2,1,1)-(5,1,1)=\langle -3,0,0\rangle .$$
Now, the normal vector $ n $ is given by
$$\overrightarrow{PQ} \times \overrightarrow{PR}=
\left|\begin{array}{rrr}{i} & {j} & {k} \\ {-4} & {0} & {1} \\ {-3} & {0} & {0}\end{array}\right|=i\left|\begin{array}{rrr} {0} & {1} \\ {0} & {0}\end{array}\right|-j\left|\begin{array}{rrr} {-4} & {1} \\ {-3} & {0}\end{array}\right|+k\left|\begin{array}{rrr} {-4} & {0} \\ {-3} & {0} \end{array}\right|\\
=i(0-0)-j(0+3)+k(0-0)\\
=-3j.
$$
Now, using any of the given points, the equation of the plane is given by
$$0(x-1)-3(y-1)+0(z-1)=0\Longrightarrow -3y+3=0$$
Hence the equation is given by
$$ y=1.$$
