Question

Equation of a plane through the line of intersection of planes 2x + 3y – 4z = 1 and 3x – y + z + 2 = 0 and parallel to

12x – y = 0 is (2x + 3y – 4z –1) + l (3x – y + z + 2) = 0 then l is –

• A. – 4
• B. $\frac { 1 } { 4 }$
• C. 4
• D. –$\frac { 1 } { 4 }$

Answer 1

Answer: (C)

4

Answer 2

(2 + 3l)x + (3 – l) y + (–4 + l) z = 1 – 2l

It is parallel to 12x – y = 0 if

$\frac { 2 + 3 \lambda } { 12 }$= $\frac { 3 - \lambda } { - 1 }$= $\frac { \lambda - 4 } { 0 }$Ž l = 4