# I need help please on this question A plane passes through the point (1,1,1) and is perpendicular to each of the planes 3x−2y+3z+6=0 and 6x−2y−3z−6=0. Find its equation.

A plane passes through the point (1,1,1) and is perpendicular to each of the planes
3x−2y+3z+6=0 and 6x−2y−3z−6=0. Find its equation. The problem is I don't have an idea of the concept. All I know is that the normal of first equation is (3,−2,3) and that of the second is (6,−2,−3).
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Dayana Powers
Here’s a hint:
the normal of the third plane is perpendicular to both normals of the two given planes.
Use the cross product.
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kasibug1v
So if T(x,y,z) is in this plane and A(1,1,1) then
$\stackrel{\to }{AT}=\left(x-1,y-1,z-1\right)=m\left(3,-2,3\right)+n\left(6,-2,-3\right)$
for some scalars m,n. Eliminate the scalars and you are done.