Question

The equation of the plane passing through the lines $\frac{x-4}{1}=\frac{y-3}{1}=\frac{z-2}{2}$ and $\frac{x-3}{1}=\frac{y-2}{-4}=\frac{z}{5}$ is

• A. $11x-y-3z=35$
• B. $11x+y-3z=35$
• C. $11x-y+3z=35$
• D. None of these

Answer 1

The correct option is D

None of these

Here, the required plane is
$a(x-4)+b(y-3)+c(z-2)=0$
Since, the normal to the plane is perpendicular to both the lines, we have
$a+b+2c=0anda-4b+5c=0$
On solving, we have
$\frac{a}{5+8}=\frac{b}{2-5}=\frac{c}{-4-1}=k\frac{a}{13}=\frac{b}{-3}=\frac{c}{-5}=k$
Therefore, the required equation of plane is $-13x+3y+5z+33=0$