Let the given points be P(2,1,2),Q(3,-8,6), and R(-2,-3,1)

\(\vec{PQ}=\langle3-2,-8-1,6-2\rangle=\langle1,-9,4\rangle\)

\(\vec{PR}=\langle-2-2,-3-1,1-2\rangle=\langle-4,-4,-1\rangle\)

\(\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\1&-9&4\\-4&-4&-1\end{vmatrix}=(9+16)i+(-16+1)j+(-4-36)k\)

\(=25i-15j-40k\)

Therefore, the normal vector to the plane is

\(\vec{n}=\langle25,-15,-40\rangle\)

Since, the plane passes through all the three points we can choose any point to find its equation. So, the equation of the plane through the point P(2,1,2) with normal vector \(\vec{n}=\langle25,-15,-40\rangle\) is

\(25(x-2)-15(y-1)-40(z-2)=0\)

\(\Rightarrow 25x-50-15y+15-40z+80=0\)

\(\Rightarrow 25x-15y-40z+45=0\)