Question

The equation of a plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point (0,7, - 7) is

• A. $x+y+z=0$
• B. $x+2y+z=21$
• C. $3x-2y+5z+35=0$
• D. $3x+2y+5z+21=0$

Answer 1

Answer: (A)

The equation of a plane containing the line
$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ is a $(x+1)+b(y-3)+c(z+2)=0$ where
$-3a+2b+c=0...\left(A\right)$
This passes through $\left(0,7,-7\right)$
$\therefore a\left(0+1\right)+b\left(7-3\right)+c\left(-7+2\right)=0$
$\Rightarrow a+4b-5c=0...\left(B\right)$
On solving equation $\left(A\right)$ and $\left(B\right)$ we get
$a=1,b=1,c=1$
$\therefore$ Required plane is
$x+1+y-3+z+2=0$
$\Rightarrow x+y+z=0$