Find the line s through A, parallel to the plane and perpendicular to r.

Let be the plane x-2y+3z-1=0 and the line r
x=2t-6
y=3t-10
z=-t+3
and the pointA (1,1,1). Find the line s through A, parallel to the plane and perpendicular to r.
Thanks a lot

The line r has direction vector <2, 3, -1> (coefficients from t).
The plane has normal vector <1, -2, 3> (coefficients from x, y, z).

A line parallel to the plane must be perpendicular to <1, -2, 3>.
Moreover, we know that this desired line s is given to be perpendicular to <2, 3, -1>.

So, the direction vector for 's' is given by the cross product
<1, -2, 3> x <2, 3, 1> = <-11, 5, 7>.

So, thee equation of 's' is given by
x = -11t + 1, y = 5t + 1, z = 7t + 1.

I hope this helps!