To find the equation of the plane we need to find its normal vector.

Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)

Note that \(\vec{PQ}\) and \(\vec{PR}\) are vectors in the plane

Hence, their cross product will give us normal vector to the plane

\(\vec{PQ}=(3,-8,6)-(2,1,2)=(3-2,-8-1,6-2)=(1,-9,4)\)

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

\(n=\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\1&-9&4\\-4&-4&-1 \end{vmatrix}=(25,-15,-40)\)

Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is

\(l(x-a)+m(y-b)+n(z-c)=0\)

We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)

Equation of the plane is

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

Divide throughout by 5

\(5(x-2)-3\cdot(y-1)-8\cdot(z-2)=0\)

\(5x-3y-8z+9=0\)

Result:

\(5x-3y-8z+9=0\)

Let us call the given points as: P(2,1,2), Q(3,-8,6), R(-2,-3,1)

Note that \(\vec{PQ}\) and \(\vec{PR}\) are vectors in the plane

Hence, their cross product will give us normal vector to the plane

\(\vec{PQ}=(3,-8,6)-(2,1,2)=(3-2,-8-1,6-2)=(1,-9,4)\)

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

\(n=\vec{PQ}\times\vec{PR}=\begin{vmatrix}i&j&k\\1&-9&4\\-4&-4&-1 \end{vmatrix}=(25,-15,-40)\)

Equation of a plane passing through the point (a,b,c) and having normal vector (l,m,n) is

\(l(x-a)+m(y-b)+n(z-c)=0\)

We found the normal vector in the previous cell. For a point on the plane, we can choose any of the three given points, I will choose P(2,1,2)

Equation of the plane is

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

Divide throughout by 5

\(5(x-2)-3\cdot(y-1)-8\cdot(z-2)=0\)

\(5x-3y-8z+9=0\)

Result:

\(5x-3y-8z+9=0\)