If for some \(\alpha \in \mathbb{R}\), the lines \(\mathrm{L}_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}\) and \(\mathrm{L}_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}\) are coplanar, then the line \(\mathrm{L}_2\) passes through the point

Here \(x_1, y_1, \mathrm{z}_1=-1,2,1\) and \(x_2, y_2, \mathrm{z}_2=-2,-1,-1\) \(a_1, b_1, c_1=2,-1,1\) and \(a_2, b_2, c_2=\alpha, 5-\alpha, 1\) Since the given lines are coplanar, we get
\(\begin{aligned} & \left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc}-2+1 & -1-2 & -1-1 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1\end{array}\right|=0\end{aligned}\)
\(\begin{aligned}
\Rightarrow & (-1)(-1-5+\alpha)-(-3)(2-\alpha) \\
& +(-2)(10-2 \alpha+\alpha)=0 \\
\Rightarrow & 6-\alpha+6-3 \alpha-20+2 \alpha=0 \\
\Rightarrow & \alpha=-4 \\
\Rightarrow & L_2: \frac{x+2}{-4}=\frac{y+1}{9}=\frac{z+1}{1}
\end{aligned}\)
\(\therefore \quad\) Only option (B) satisfies the equation of line \(\mathrm{L}_2\).