If for some m in mathbb{R} the lines L_1 : fr | vedclass

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(B) 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

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$(7, -2, 5)$

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(B) 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 $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.Here, $(x_1, y_1, z_1) = (-1, m, 1)$ and $(x_2, y_2, z_2) = (-2, -1, -1)$.So, $x_2-x_1 = -1$, $y_2-y_1 = -1-m$, $z_2-z_1 = -2$.The determinant is $\begin{vmatrix} -1 & -(1+m) & -2 \ m & -1 & 1 \ -4 & 9 & 1 \end{vmatrix} = 0$.Expanding along the first row: $-1(-1-9) + (1+m)(m+4) - 2(9m-4) = 0$.$10 + (m^2 + 5m + 4) - 18m + 8 = 0$.$m^2 - 13m + 22 = 0$.$(m-11)(m-2) = 0$, so $m=2$ or $m=11$.If $m=2$, $L_1 : \frac{x+1}{2} = \frac{y-2}{-1} = \frac{z-1}{1}$. Point $(7, -2, 5)$ satisfies this: $\frac{7+1}{2} = 4$, $\frac{-2-2}{-1} = 4$, $\frac{5-1}{1} = 4$. Thus, $L_1$ passes through $(7, -2, 5)$.

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

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