Like plugging in a point in a 2D line you are doing the same here with one other step!
You need to make sure the direction vector you are investigating is going in the proper direction in relation to the plane given!
You simply plug in the line given to you in this problem to see if it works for the equation of the given 3D plane!
Solve the system of equations (in parametric form)!!
So we have:
y= -2 - (s) - (2t)= 1
z= 5+(0s) - (t)= -2 (where (2,1,-2) is the point d given)
Multiply 6 to equation z to cancel out the t terms in equation z and equation x when you combine the two equations
you get 34 +(2(s))= -10 and solve for s !
you get s = -22 and then plug into one of the original parametric equation you made from the given!
solve for t in equation x you get :
4 + (2(-22)) + (6(t)) = 2 isolate t you get:
6(t)=42 and divide by 6 to get:
So this direction vector does exist on the given plane!
Therefore this direction vector does lie on plane given!!