The equation of any plane through (2, 1, 0) is
A(x – 2) + B(– 1) + C(z – 0) = 0    ...(1)
∴ it passes through (3, 2, 2)
∴ A(3 – 2) + B(2 – 1) + C(2 – 0) = 0
∴ A + B + 2C = 0    ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, 1.
∴  (2) (A) + (3) (B) + (1) (C) = 0
∴  2A + 3B + C = 0    ...(3)
Solving (2) and (3), we get,
                        


∴ A = 5 k, B = – 3 k, C = – k
Putting these values of A, B, C in (1), we get,
5 k (x – 2) – 3 k (y – 1) – k (z – 0) = 0
or 5 (x – 2) – 3(y – 1) – z = 0
or 5 x – 10 – 3 y + 3 – z = 0
or 5 x – 3 y – z – 7 = 0
which is required equation of plane.