NDA Math Three Dimensional Geometry

Since we have the plane 2x+3y−z=7,
we know that the vector is [2,3,−1] perpendicular to the plane
With the point (1,−2,3) we get the parametric equations for the line
x=1+2t,y=−2+3t,z=3−t
The line equation can be found by solving for t
t=

(x−1)

2

=

(y+2)

3

=−

(z−3)

1

For intersection point
⇒2+4t−6+9t−3+t=7
⇒t=1
Therefore, the point is
(1+2,−2+3,3−1)≡(3,1,2)