We know that two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Let us verify this option.

The given lines are $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$.

$=\left|\begin{array}{ccc}-1-\left(-3\right)& 2-1& 5-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|$

$=\left|\begin{array}{ccc}2& 1& 0\\ -3& 1& 5\\ -1& 2& 5\end{array}\right|$

$=2\left(5-10\right)-\left(1\right)\left(-15-\left(-5\right)\right)+0$

$=2\left(-5\right)+\left(10\right)=0$

Therefore, this is the correct option.