The line, that is coplanar to the line $$\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$$, is

Given: Line $$\frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$$ passes through $$(-3,1,5)$$ with direction $$(-3,1,5)$$.

We need a coplanar line from the options. Two lines are coplanar if the determinant of the matrix formed by the direction vectors and the connecting vector equals zero.

For option 2 (assuming it's distinct from option 1): Line through $$(-1,2,5)$$ with direction $$(-1,2,5)$$.

Connection vector: $$(-1+3, 2-1, 5-5) = (2,1,0)$$.

$$\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = 2(5-10) - 1(-15+5) + 0 = -10 + 10 = 0$$ ✓ Coplanar.

The correct answer is Option B.