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)

$x+y+z=0$

The equation of plane containing the line

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

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

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

This passes through $(0,7,-7)$.

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

From equations (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 $x+y+z=0$