Explanation

To find the value of $26 \alpha(\mathrm{PB})^2$, we proceed as follows:

Intersection Point $ B $ of $\mathrm{L}_1$ and $\mathrm{L}_2$

We are given the equations of the lines:

$ \mathrm{L}_1 : \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} $

$ \mathrm{L}_2 : \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha} $

For $\mathrm{L}_1$, $ z+1 = 0 $, which implies $ z = -1 $.

For $\mathrm{L}_2$, since $ y = 0 $, the direction ratios can be matched using the parameter $\mu$:

$ \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} $

Setting the coordinates equal for intersection:

$ \begin{aligned} & 3\lambda + 1 = 2\mu + 2, \\ & -\lambda + 1 = 0, \\ & -1 = \alpha \mu - 4. \end{aligned} $

From $-\lambda + 1 = 0$, we find $\lambda = 1$.

Substituting $\lambda = 1$ into $3\lambda + 1 = 2\mu + 2$ gives:

$ 3(1) + 1 = 2\mu + 2 \implies \mu = 1. $

Also, substituting $\mu = 1$ into $-1 = \alpha \mu - 4$ gives:

$ -1 = \alpha(1) - 4 \implies \alpha = 3. $

So, the intersection point $B$ is:

$ B(4, 0, -1). $

Foot of Perpendicular from $A$ to $\mathrm{L}_2$

The point $P$ on $\mathrm{L}_2$ is given by:

$ P(2\delta + 2, 0, 3\delta - 4). $

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:

$ \begin{aligned} (2\delta + 1)\cdot 2 + (-1)\cdot 0 + (3\delta - 3)\cdot 3 &= 0. \end{aligned} $

Simplifying gives:

$ 4\delta + 2 + 9\delta - 9 = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}. $

So, the coordinates of point $P$ are:

$ P\left(\frac{40}{13}, 0, \frac{-31}{13}\right). $

Distance $\mathrm{PB}$ and Calculation

The vector $\overrightarrow{PB}$ is:

$ \overrightarrow{PB} = \left(4 - \frac{40}{13}, 0 - 0, -1 - \left(\frac{-31}{13}\right)\right). $

Calculating the components:

$ \overrightarrow{PB} = \left(\frac{12}{13}, 0, \frac{18}{13}\right). $

The square of the distance $(PB)^2$ is:

$ \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}. $

Thus, $26 \alpha (PB)^2$ is:

$ 26 \times 3 \times \frac{468}{169} = 216. $

So, the final value is 216.