(i) 2x + y + z – 3 + λ(x – y – z – 4) = 0
⇒ (2 + λ)x + (1 – λ)y + (1 – λ)z – 3 – 4λ = 0
Vector equation is .
\(\bar{r}\).(2 + λ)i + (1 – λ)y + (1 – λ)k – (3 + 4)λ = 0
(ii) Since passing through (1,2, – 1) we have;
⇒ (2 + λ)1 + (1 – λ)2 + (1 – λ)( -1) – 3 – 4λ = 0
⇒ 2 + λ + 2 – 2λ – 1 + 1 – 3 – 4λ = 0
⇒ 0 – 4λ – 0
⇒ λ = 0
\(\bar{r}\).(2i + j + k) = 3 is required plane. Since the point (1, 2, -1) is a point on the first plane