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Answer:
Step-by-step explanation:
Given that the plane passes through (1,3,3)
Also the plane contains the line x(t)=3t,y(t)=2t,z(t)=4+3t.
This means all points lying in this line will also lie in the plane.
Find out two points on this line.
First point: Let t =0. Point is (0,0,4)
Next point : Let t = 1: Point is (3,2,7)
Now we have 3 non collinear points (0,0,4) (3,2,7) and (1,3,3) lying on the plane.
Equation of the plane is
Simplify to get
x(-2-9)-y(-3-3)+(z-4)(9-2)=0
i.e -11x+6y+7z-28 =0
11x-6y-7z+28 =0 is the equation of the plane.