Question

Equation of plane containing the line $\frac{x-1}{-3}=\frac{y-3}{2}=\frac{z-2}{1}$ and the point $(0,-3,4)$ is

• A. $x+y+2z-5=0$
• B. $2x-y+3z-15=0$
• C. $2x+y+4z=13$
• D. $x+y-4z=13$

Answer 1

Answer: (C)

Equation of plane containing the line
$a(x-1)+b(y-3)+c(z-2)=0$ and $-3a+2b+c=0$
Using $(0,-3,4)$, we have
$a(-1)+b(-6)+c(2)=0$
or $a+6b-2c=0$
and $3a-2b-c=0$
$\therefore \frac{a}{10}=\frac{-b}{-5}=\frac{c}{20}$
or $\frac{a}{2}=\frac{b}{1}=\frac{c}{4}$
$\therefore$ Required equation of plane is
$2(x-1)+y-3+4(z-2)=0$
$\Rightarrow 2x+y+4z=13$