Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes
x + 2y + 3z = 5 and 3x + 3y + z = 0.

We know, the plane passing through (x₁, y₁, z₁) and having direction ratios of normal as A, B and C have equation

A(x – x₁) + B(y – y₁) + C(z – z₁) = 0

Now, required plane passes through (-1, 3, 2) and let A, B and C be the direction ratios of normal of given plane

Then, equation of the line is

A(x + 1) + B(y – 3) + C(z – 2) = 0

Since the line is perpendicular to the planes x + 2y + 3z = 5 and 3x + 3y + z = 0,normal to plane would be perpendicular to normal of these two planes

Also, we know that

is perpendicular to both and

Therefore, we have to find the cross product of normal of plane

x + 2y + 3z = 5 [Normal ]

3x + 3y + z = 0 [Normal ]

Hence, required normal

Hence, direction ratios are -7, 8, -3

And equation of plane:

-7(x + 1) + 8(y – 3) – 3(z – 2) = 0

⇒ -7x – 7 + 8y – 24 – 3z + 6 = 0

⇒ 7x – 8y + 3z + 25 = 0