Find the scalar equation of the plane containing the parallel lines:

r=(2,2,-1)+s(3,-3,1) and r=(1,2,-3)+t(3,-3,1)

Can someone check my work for accurateness? I don't have the answer to this question and would like to know if it is correct. Thanks for your help

P(2,2,-1)

Q(1,2,-3)

d3=PQ=(-1,0,-2)

n=d1 x d3

n=(6,5,-3)

6x+5y-3z+D=0

6(1)+5(2)-3(-3)+D=0

D=-25

Final answer:

6x+5y-3z-25=0