QJEEAdv10Ma64f2: JEE Advanced 2010 Mathematics - Class 12 JEEAdv10Ma64f2 Solution: Equation of the plane containing the straight line and…

Plane 1 : $ax + by + cz = 0$ contains line ${x \over 2} = {y \over 3} = {z \over 4}$ . Therefore,

$2a + 3b + 4c = 0$ ..... (1)

Plane 2 : $a'x + b'y + c'z = 0$ is perpendicular to plane containing line ${x \over 3} = {y \over 4} = {z \over 2}$ and ${x \over 4} = {y \over 2} = {z \over 3}$ .

Hence, $3a' + 4b' + 2c' = 0$ and $4a' + 2b' + 3c' = 0$ .

$\Rightarrow {{a'} \over {12 - 4}} = {{b'} \over {8 - 9}} = {{c'} \over {6 - 16}}$

$\Rightarrow 8a - b - 10c = 0$ .... (2)

From Eq. (1) and (2), we get

${a \over { - 30 + 4}} = {b \over {32 + 20}} = {c \over { - 2 - 24}}$

$\Rightarrow$ Equation of plane is $x - 2y + z = 0$