Any plane through (1, 1, 0) isa(x – 1) + b(y – 1) + cz = 0 ...(1)Since the plane passes through the points (1, 2, 1) and (–2, 2, – 1),∴ a(1 – 1) + b(2 – 1) + c(1) = 0 and a(–2 – 1) + b(2 – 1) + c(–1) = 0⇒ b + c = 0 ...(2)and –3a + b – c = 0 ...(3)Solving (2) and (3), $\frac{a}{-1-1}=\frac{b}{-3+0}=\frac{c}{0+3}$$\Rightarrow \frac{a}{-2}=\frac{b}{-3}=\frac{c}{3}=k\,\text{(say)}$, where k ≠ 0∴ a = –2k, b = –3k, c = 3k.Putting these values of a, b, c in (1), we get–2k(x – 1) – 3k(y – 1) + 3k z = 0 ⇒ – 2(x – 1) – 3 (y – 1) + 3z = 0 [since k ≠ 0]⇒ – 2x + 2 – 3y + 3 + 3z = 0⇒ 2x + 3y – 3z = 5, which is the required equation.