Question

The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point :

(1) (0, 6, –2) 

(2) (–2, 0, 1) 

(3) (0, –6, 2) 

(4) (2, 0, –1)

Answer 1

Answer is (2) (–2, 0, 1)

Two points on the line (L say) \(\frac{x}{3}=\frac{y}{2}\) ,z = 1 are (0, 0, 1) & (3, 2, 1)

So dr's of the line is < 3, 2, 0 >

Line passing through (1, 2, 1), parallel to L and coplanar with given Plane is

\(\hat i+\hat 2j +\hat k + t(\hat 3i +\hat 2j), \) t ∈ R (-2,0,1)

satisfies the line (for t = –1)

⇒ (–2, 0, 1) lies on given plane. 

Answer of the question is (2) We can check other options by finding eqution of plane