Question

The equation of the plane containing the line

$\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point (0, 7, –7) is

• A. $x + y + z = 1$
• B. $x + y + z = 2$
• C. $x + y + z = 0$
• D. None of these

Answer 1

Answer: (C)

$x + y + z = 0$

Answer 2

The equation of plane containing the line

$\frac { x + 1 } { - 3 } = \frac { y - 3 } { 2 } = \frac { z + 2 } { 1 }$ is

$a ( x + 1 ) + b ( y - 3 ) + c ( z + 2 ) = 0$ .....(i)

where $- 3 a + 2 b + c = 0$ .....(ii)

This passes through (0, 7, –7)

$\therefore$ $a + 4 b - 5 c = 0$ ..…(iii)

From (ii) and (iii),$\frac { a } { - 14 } = \frac { b } { - 14 } = \frac { c } { - 14 }$ or $\frac { a } { 1 } = \frac { b } { 1 } = \frac { c } { 1 }$

Thus, the required plane is $x + y + z = 0$.