Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26 \alpha(\mathrm{~PB})^2$ is _________ .

<p><strong>Explanation</strong></p> <p>To find the value of $26 \alpha(\mathrm{PB})^2$, we proceed as follows:</p> <p><strong>Intersection Point $ B $ of $\mathrm{L}_1$ and $\mathrm{L}_2$</strong></p> <p>We are given the equations of the lines:</p> <p>$ \mathrm{L}_1 : \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} $</p> <p>$ \mathrm{L}_2 : \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha} $</p> <p>For $\mathrm{L}_1$, $ z+1 = 0 $, which implies $ z = -1 $.</p> <p>For $\mathrm{L}_2$, since $ y = 0 $, the direction ratios can be matched using the parameter $\mu$:</p> <p>$ \begin{aligned} & x = 3\lambda + 1, \quad y = -\lambda + 1, \quad z = -1, \\ & x = 2\mu + 2, \quad y = 0, \quad z = \alpha \mu - 4. \end{aligned} $</p> <p>Setting the coordinates equal for intersection:</p> <p>$ \begin{aligned} & 3\lambda + 1 = 2\mu + 2, \\ & -\lambda + 1 = 0, \\ & -1 = \alpha \mu - 4. \end{aligned} $</p> <p>From $-\lambda + 1 = 0$, we find $\lambda = 1$.</p> <p>Substituting $\lambda = 1$ into $3\lambda + 1 = 2\mu + 2$ gives:</p> <p>$ 3(1) + 1 = 2\mu + 2 \implies \mu = 1. $</p> <p>Also, substituting $\mu = 1$ into $-1 = \alpha \mu - 4$ gives:</p> <p>$ -1 = \alpha(1) - 4 \implies \alpha = 3. $</p> <p>So, the intersection point $B$ is:</p> <p>$ B(4, 0, -1). $</p> <p><strong>Foot of Perpendicular from $A$ to $\mathrm{L}_2$</strong></p> <p>The point $P$ on $\mathrm{L}_2$ is given by:</p> <p>$ P(2\delta + 2, 0, 3\delta - 4). $</p> <p>For vector $\overrightarrow{AP} = (2\delta + 1, -1, 3\delta - 3)$, since $AP$ is perpendicular to $\mathrm{L}_2$, the dot product should be zero:</p> <p>$ \begin{aligned} (2\delta + 1)\cdot 2 + (-1)\cdot 0 + (3\delta - 3)\cdot 3 &= 0. \end{aligned} $</p> <p>Simplifying gives:</p> <p>$ 4\delta + 2 + 9\delta - 9 = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}. $</p> <p>So, the coordinates of point $P$ are:</p> <p>$ P\left(\frac{40}{13}, 0, \frac{-31}{13}\right). $</p> <p><strong>Distance $\mathrm{PB}$ and Calculation</strong></p> <p>The vector $\overrightarrow{PB}$ is:</p> <p>$ \overrightarrow{PB} = \left(4 - \frac{40}{13}, 0 - 0, -1 - \left(\frac{-31}{13}\right)\right). $</p> <p>Calculating the components:</p> <p>$ \overrightarrow{PB} = \left(\frac{12}{13}, 0, \frac{18}{13}\right). $</p> <p>The square of the distance $(PB)^2$ is:</p> <p>$ \left(\frac{12}{13}\right)^2 + \left(0\right)^2 + \left(\frac{18}{13}\right)^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}. $</p> <p>Thus, $26 \alpha (PB)^2$ is:</p> <p>$ 26 \times 3 \times \frac{468}{169} = 216. $ </p> <p>So, the final value is 216.</p>