Question

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$is

• A. (0, 1, –2)
• B. (1, 2, 3)
• C. (–1, 9, –25)
• D. $\left( \frac{- 1}{11},\frac{9}{11}\frac{- 25}{11} \right)$

Answer 1

Answer: (D)

$\left( \frac{- 1}{11},\frac{9}{11}\frac{- 25}{11} \right)$

Answer 2

$\frac { x } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 2 } { 3 } = r$, (say)

So, $x = r$, $y = 2 r + 1$, $z = 3 r - 2$

$\therefore 2 r + 3 ( 2 r + 1 ) + ( 3 r - 2 ) = 0 \Rightarrow r = \frac { - 1 } { 11 }$

Hence, $x = \frac { - 1 } { 11 } , y = \frac { 9 } { 11 } , z = - \frac { 25 } { 11 }$.