If the two lines $\frac{x-1}{2}=\frac{y-3}{4}=-z$ and $\frac{x-4}{3}=\frac{y-1}{-2}=z-1$ are contained in a plane then the equation of the plane is:

$2 x-5 y-16 z+13=0$

The correct answer is Option (1) → $2 x-5 y-16 z+13=0$

$l_1:\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-0}{-1}$

$l_2:\frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1}$

$\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\2&4&-1\\3&-2&1\end{vmatrix}$

$\vec n=2\hat i-5\hat j-16\hat k$

let $\vec a=\hat i+3\hat j+0\hat k$

on plane from $l_1$

$⇒(\vec r-\vec a).\vec n = 0$

$⇒(x\hat i+y\hat j+z\hat k-\hat i-3\hat j)(2\hat i-5\hat j-16\hat k)=0$

$⇒2x-5y-16z-2+15=0$

$⇒2x-5y-16z+13=0$