How to find an equation of the plane that contains the line and is perpendicular to the plane

line: (x+2)/3=(y-4)/2=3-z
plane: x-2y+z=5

The line is:

(x+2)/3 = (y-4)/2 = 3-z = t

Or in parametric form

x = 3t - 2, y = 2t + 4, z = (-1)t + 3

So the direction of the line is <3,2,-1>.

The normal direction to the plane is <1,-2,1> hence the cross product

<3,2,-1> x <1,-2,1> = <0,-4,-8>

is perpendicular to both. The plane must therefore have the form

(0)x - 4y - 8z = C

Setting t=0 in the equation for the line gives (-2,4,3) as a point on the line, hence

(0)(-2) - (4)(4) - (8)(3) = -40

So factoring out the common factor of four we get the plane

y + 2z = 10