Finding an equation of the plane (Linear Algebra)

[SmellyGoomba]

Homework Statement

Find an equation of the plane that has y-intercept -5 and is parallel to the plane containing the points P(3, -1, 2), Q(0, 2, 1) and R(5, 2, 0)

Homework Equations

ax + by + cz + d = 0

The Attempt at a Solution

I got two directional vectors
u = PQ = (-3, 3, -1)
v = PR = (2, 3, -2)

n = u x v = (-3, -8, -15) which is also equal to (3, 8, 15) because it's a nonzero multiple of n

So putting P and the norm together, I get
3(x-3) + 8(y+1) + 15(z-2) = 0
3x + 8y + 15z - 31 = 0

Now I'm stuck. I have no idea what to do with y-intercept of -5. A push in the right direction would be nice.

 

[slider142]

The point P is not in your plane. It is in a parallel plane. The y-intercept (0, -5, 0) is the only point that is presented as being in your plane.