Question

Determine the vector and parametric equations of the plane that passes through the points Q(-3/2 , 0 , 0) , R(0,-1,0) and S (0,0,3)

b) Determine if the point P(1,5,6) is a point on this plane

Answer 1

you will need two direction vectors on the plane
how about vectors RS and RQ ?

vector RS = (0, 1, 3)
vector RQ = (-3/2, 0, -3) or more civil-like : (1, 0, 2)

vector equation:
(x,y,z) = (0, -1, 0) + s(0, 1, 3) + t(1, 0 , 2)

parametric form:
x = 0 + t
y = -1 + s
z = 3s + 2t

b) is (1,5,6) on it?
1 = 0+t , t = 1
5 = -1 + s, s = 6

then 6 = 3(6) + 2(1), which is false, so (1,5,6) is NOT on the plane

other method:
it is easy to find the cross-product of our two direction vectors to be (2,3,-1)
so the plane equation is 2x + 3y - z = c
but (0,0,3) lies on it, so 0 + 0 - 3 = c

plane equation is 2x + 3y - z = -3
plug (1,5,6) into it ...
LS = 2(1) + 3(5) - 6 ≠ -3
so the point is NOT on the plane