Plane:
Generally: $ \vec r = \vec a + t \vec u + s \vec v$;
This plane passes through the point
specified by $\vec a$.
The direction vectors $ \vec u $ and $\vec v$,
starting from this point, vector addition, span the plane.
$\vec a = (0,0,0), \vec u = (1,2,3)$ , and
$\vec v= (-2,1,1).$
For the problem (solution)
the normal vector $ \vec n $, perpendicular
to both direction vectors is required.
Cross product of vectors:
$\vec n = \vec u × \vec v = (-1,-7,5)$;
Plane:
$\vec n \cdot ( \vec r - \vec a) = 0$,
$(-1,-7,5) \cdot (x,y,z) =0$.
Finally: $ -x -7y +5z = 0.$
Helps?
Best Answer
A point on the normal to the plane passing through the origin has coordinates $(3t,-2t,-t)$ [because $(3,-2,-1)$ is normal to the plane] and if it is on the plane $3x-2y-z=-4$ we have $9t+4t+t=-4$ so $t=-\dfrac{2}{7}$ and the point is $(-6/7,4/7,2/7)$.