Determine the Scalar equation of the plane that passes through the point (2, -4, -3), and contains the vectors a = [-3, 5, -1] and b = [2, -4, 6]

Answer 1

We need the normal to both a and b which would be the cross-product a X b
I assume you know how to find the cross-product and that you got <13,8,1> in reduced form

So the equation of the plane is 13x + 8y + z = c
but (2,-4,-3) lies on it, so
26 - 32 - 3 = c = -9

Plane: 13x + 8y + z = -9

better check my arithmetic