Equation of plane from normal A line is given by
    $x = 2 + 5t ,  y = 1 + 2t ,  z = 3 − 3t$
Define an equation in normal form for the plane which is perpendicular to the line and which intersects the point $(2, −3, 4)$.
When I look at the answer it says that the normal to the plane is
$ n = [5; 2; -3]$
which is also the direction vector of the line, which I understand. Then we get to the part that confuses me. Here they claim that
$`5(x − 2) + 2(y + 3) − 3(z − 4) = 0`$
or
$`5x + 2y − 3z + 8 = 0`$
is the equation of the plane we seek, but I don't follow. What's a more detailed way of describing this approach?

hint
If $(a,b,c)$ is a normal vector to a plane, then the cartesian equation of this plane will be of the form
$$ax+by+cz=d.$$
if the plane contains the point $(x_0,y_0,z_0)$, the equation becomes
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$
with $d=ax_0+by_0+cz_0$.

parmetric equations of the plane

the vectors director of the plane can be taken as
$u=(-b,a,0)\perp(a,b,c)$ and
$v=(-c,0,a)\perp(a,b,c).$
