The straight line joining the points
\[
(2,-2,1)
\quad\text{and}\quad
(-2,1,-1)
\]
is perpendicular to the plane
\[
\pi_1
\]
which passes through the point
\[
(1,1,1)
\]
and the equation of \(\pi_1\) is
\[
ax+by+cz+d=0.
\]
If
\[
\pi_2
\]
is the plane passing through the point
\[
(1,2,3)
\]
whose normal vector is
\[
(a,b,c),
\]
then the equation of \(\pi_2\) is
Show Hint
If a line is perpendicular to a plane, then the direction vector of the line is the normal vector of the plane.
The equation of a plane through \((x_1,y_1,z_1)\) with normal vector \((a,b,c)\) is
\[
\boxed{
a(x-x_1)+b(y-y_1)+c(z-z_1)=0.
}
\]