Question

The equation of the plane containing the two lines $\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z}{3}$ and $\frac{x}{2}=\frac{y-2}{-1}=\frac{z+1}{3}$ is

• A. $8x+y-5z-7=0$
• B. $8x+y+5z-7=0$
• C. $8x-y-5z-7=0$
• D. None of these

Answer 1

The correct option is A $8x+y-5z-7=0$

Equation of any plane through the first line is

$a(x-1)+b(y+1)+cz=0$ ...$\left(1\right)$

where $2a-b+3c=0$ ...$\left(2\right)$

It will pass through the second line if $(0,2,-1),$ a point on the second line lies on it

i.e., if $-a+3b-c=0$ ....$\left(3\right)$

Solving $\left(2\right)$ and $\left(3\right)$, we get

$\frac{a}{-8}=\frac{b}{-1}=\frac{c}{5}\Rightarrow \frac{a}{8}=\frac{b}{1}=\frac{c}{-5}$

Hence equation of required plane is

$8(x-1)+(y+1)-5z=0\Rightarrow 8x+y-5z-7=0$