Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar.

(N/A) 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 $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
From the given equations:
$(x_1, y_1, z_1) = (-3, 1, 5)$ and $(a_1, b_1, c_1) = (-3, 1, 5)$.
$(x_2, y_2, z_2) = (-1, 2, 5)$ and $(a_2, b_2, c_2) = (-1, 2, 5)$.
Now,calculate the determinant:
$\begin{vmatrix} -1-(-3) & 2-1 & 5-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$.
Expanding along the first row:
$= 2(1 \times 5 - 5 \times 2) - 1(-3 \times 5 - 5 \times -1) + 0(-3 \times 2 - 1 \times -1)$
$= 2(5 - 10) - 1(-15 + 5) + 0$
$= 2(-5) - 1(-10) = -10 + 10 = 0$.
Since the determinant is $0$,the lines are coplanar.