If for some $m \in \mathbb{R}$ the lines $L_1 : \dfrac{x+1}{m} = \dfrac{y-m}{-1} = \dfrac{z-1}{1}$ and $L_2 : \dfrac{x+2}{-4} = \dfrac{y+1}{9} = \dfrac{z+1}{1}$ are coplanar, then line $L_1$ passes through the point

For the lines $L_1$ and $L_2$ to be coplanar, the shortest distance between them must be zero. The condition for coplanarity of two lines is given by:

$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$

From the given equations, the lines pass through the points $(-1, m, 1)$ and $(-2, -1, -1)$ respectively. Their direction ratios are $(m, -1, 1)$ and $(-4, 9, 1)$.

Substituting these values into the determinant:

$\begin{vmatrix} -2 - (-1) & -1 - m & -1 - 1 \\ m & -1 & 1 \\ -4 & 9 & 1 \end{vmatrix} = 0$

$\begin{vmatrix} -1 & -1-m & -2 \\ m & -1 & 1 \\ -4 & 9 & 1 \end{vmatrix} = 0$

Expanding the determinant along the first row:

$-1(-1 - 9) - (-1 - m)(m - (-4)) - 2(9m - 4) = 0$

$10 + (1 + m)(m + 4) - 18m + 8 = 0$

$10 + m^2 + 5m + 4 - 18m + 8 = 0$

$m^2 - 13m + 22 = 0$

$(m - 2)(m - 11) = 0$

This gives $m = 2$ or $m = 11$.

For $m = 2$, the equation of line $L_1$ becomes:

$\dfrac{x+1}{2} = \dfrac{y-2}{-1} = \dfrac{z-1}{1}$

Checking the given options by substituting the coordinates into the equation of $L_1$ for $m = 2$:

For the point $(7, -2, 5)$:

$\dfrac{7+1}{2} = \dfrac{8}{2} = 4$

$\dfrac{-2-2}{-1} = \dfrac{-4}{-1} = 4$

$\dfrac{5-1}{1} = \dfrac{4}{1} = 4$

Since all three parts are equal, the line $L_1$ passes through the point $(7, -2, 5)$.