Find properties of a plane given three points that determine it

Let $M$ be the plane determined by the three points $(1,1,-1)$ , $(3,3,2)$ , and $(3,-1,-2)$ . Find the following:

Denote the points by $P = (3,-1,-2)$ , $Q = (1,1,-1)$ and $R = (3,3,2)$ , then we can compute a normal vector by
$\begin{align*} N &= (P-Q) \times (P-R) \\ &= ((3,-1,-2) - (1,1,-1)) \times ((3,-1,-2) - (3,3,2)) \\ &= (2,-2,-1) \times (0,-4,-4) \\ &= (4,8,-8). \end{align*}$

Therefore, $N = (1,2,-2)$ is a normal vector to the plane.
Since $(1,2,-2)$ is normal to the plane we have a Cartesian equation of the form
$x + 2y - 2z = d.$

Then, since $(1,1,-1)$ is on the plane we have $d = 5$ . Hence, the Cartesian equation is

$x + 2y - 2z = 5.$
The distance from the origin is
$d = \frac{5}{ \lVert N \rVert} = \frac{5}{3}.$