Find the equation of the plane which passes through the points A(-1,-3,1), B(1,-2,0) and C(2,1,-1).
Calculate the shortest distance between the plane and the origin of the coordinate system.
Since A, B, and C lie in the plane, the vectors
$\displaystyle \overline{AB}=[2,1,-1]$
and $\displaystyle \overline{AC}=[3,4,-1]$ are parallel to the plane.
Therefore, $\displaystyle \overline{AB}\times \overline{AC}=\begin{vmatrix}i&j&k\\2&1&-1\\3&4&-1\end{vmatrix}$
$\displaystyle =2i+j+5k$
is normal to the plane, since it is perpendicular to $\displaystyle \overline{AB}, \;\ and \;\ \overline{AC}$.
By using the normal and the point (-1,-3,1) in the plane, we get the
point-normal form:
$\displaystyle 2(x+1)+(y+3)+5(z-1)$
$\displaystyle 2x+y+5z=0$
To find the distance between a point and a plane, use the formula:
$\displaystyle D=\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\sqrt{a^{2}+b^{2 }+c^{2}}}$