Answer
$$6x+9y+4z=19.$$
Work Step by Step
We have
$$\overrightarrow{PQ}=Q-P=(1,1,1)-(2,-1,4)=\langle -1,2,-3\rangle,$$
$$\overrightarrow{PR}=R-P=(3,1,-2)-(2,-1,4)=\langle 1,2,-6\rangle .$$
Now, the normal vector $ n $ is given by
$$\overrightarrow{PQ} \times \overrightarrow{PR}=
\left|\begin{array}{rrr}{i} & {j} & {k} \\ {-1} & {2} & {-3} \\ {1} & {2} & {-6}\end{array}\right|=i\left|\begin{array}{rrr} {2} & {-3} \\ {2} & {-6}\end{array}\right|-j\left|\begin{array}{rrr} {-1} & {-3} \\ {1} & {-6}\end{array}\right|+k\left|\begin{array}{rrr} {-1} & {2} \\ {1} & {2} \end{array}\right|\\
=i(-12+6)-j(6+3)+k(-2-2)\\
=-6i-9j-4k.
$$
Now, using any of the given points, the equation of the plane is given by
$$-6(x-1)-9(y-1)-4(z-1)=0\Longrightarrow -6x-9y-4z+19=0$$
Hence the equation is given by
$$6x+9y+4z=19.$$
