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. $9x+3y-5z+21=0$
• B. $13x-y+z+14=0$
• C. $9x-y-z-14=0$
• D. $13x-3y-5z-33=0$

Answer 1

The correct option is D $13x-3y-5z-33=0$

Given: $L_1:\frac{x-4}{1}=\frac{y-3}{1}=\frac{z-2}{2}$ and $L_2:\frac{x-3}{1}=\frac{y-2}{-4}=\frac{z}{5}$
D.R's of $L_1=\hat{i}+\hat{j}+2\hat{k}$
DR's of $L_2=\hat{i}-4\hat{j}+5\hat{k}$
Required plane also passes through $(4,3,2)$
So plane equation is given by
$\left|\begin{array}{ccc}x-4& y-3& z-2\\ 1& 1& 2\\ 1& -4& 5\end{array}\right|=0\Rightarrow 13(x-4)-3(y-3)-5(z-2)=0$
$\Rightarrow 13x-3y-5z-33=0$