This problem requires a thorough understanding of lines in 3D space, including their parametric representation, finding intersection points, determining the foot of a perpendicular from a point to a line, and calculating distances. We will systematically approach each part of the problem.

1. Key Concepts and Formulas

• Parametric Form of a Line: A line passing through a point $(x_0, y_0, z_0)$ with direction ratios $(a, b, c)$ can be expressed as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$. Any point on the line is then $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$. If a direction ratio is zero (e.g., $c=0$), it implies $z-z_0=0$, or $z=z_0$.
• Intersection of Two Lines: If two lines intersect, there exist unique parameter values (one for each line) such that the coordinates of a point on the first line are identical to the coordinates of a point on the second line.
• Foot of Perpendicular from a Point to a Line: If $P$ is the foot of the perpendicular from an external point $A$ to a line $L$, then the vector $\vec{AP}$ is perpendicular to the direction vector of the line $L$. This means their dot product is zero: $\vec{AP} \cdot \vec{d_L} = 0$.
• Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. The squared distance is $(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

2. Step-by-Step Solution

Step 1: Representing the Lines in Parametric Form

We begin by converting the given symmetric equations of the lines $L_1$ and $L_2$ into their parametric forms. This allows us to represent any point on the lines using a single variable.

For line $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$, we set each ratio equal to a parameter, say $\lambda$. The $z+1=0$ implies $z=-1$. Thus, any point on $L_1$ can be written as: $L_1: (x, y, z) = (3\lambda + 1, -\lambda + 1, -1)$ The direction vector for $L_1$ is $\vec{d_1} = \langle 3, -1, 0 \rangle$.

For line $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$, we set each ratio equal to a parameter, say $\mu$. The $y=0$ implies $y=0$. Thus, any point on $L_2$ can be written as: $L_2: (x, y, z) = (2\mu + 2, 0, \alpha\mu - 4)$ The direction vector for $L_2$ is $\vec{d_2} = \langle 2, 0, \alpha \rangle$.

Step 2: Finding the Intersection Point B and the Value of $\alpha$

The lines $L_1$ and $L_2$ intersect at point $B$. This means that for specific values of $\lambda$ and $\mu$, the coordinates of a point on $L_1$ will be equal to the coordinates of a point on $L_2$. Let $B = (x_B, y_B, z_B)$. Equating the corresponding coordinates:

1. $x$-coordinate: $3\lambda + 1 = 2\mu + 2$
2. $y$-coordinate: $-\lambda + 1 = 0$
3. $z$-coordinate: $-1 = \alpha\mu - 4$

From equation (2), we directly find $\lambda$: $-\lambda + 1 = 0 \implies \lambda = 1$

Substitute $\lambda = 1$ into equation (1): $3(1) + 1 = 2\mu + 2$ $4 = 2\mu + 2$ $2\mu = 2 \implies \mu = 1$

Now, substitute $\mu = 1$ into equation (3) to find the value of $\alpha$: $-1 = \alpha(1) - 4$ $-1 = \alpha - 4$ $\alpha = 3$

With $\lambda = 1$ (or $\mu = 1$ and $\alpha=3$), we can find the coordinates of the intersection point $B$. Using the parametric form of $L_1$: $B = (3(1) + 1, -(1) + 1, -1) = (4, 0, -1)$ We can verify this using $L_2$ with $\mu=1$ and $\alpha=3$: $B = (2(1) + 2, 0, 3(1) - 4) = (4, 0, -1)$. The coordinates match.

So, the intersection point is $B(4, 0, -1)$ and the value of $\alpha$ is $3$.

Step 3: Finding the Foot of Perpendicular P from A to $L_2$

We are given point $A(1,1,-1)$ and line $L_2$. Now we know $\alpha=3$, so the equation of $L_2$ is $\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{3}$. Let $P$ be the foot of the perpendicular from $A$ to $L_2$. Since $P$ lies on $L_2$, we can represent its coordinates using the parametric form of $L_2$ with $\alpha=3$, using a new parameter, say $\delta$: $P(2\delta + 2, 0, 3\delta - 4)$

Next, we form the vector $\vec{AP}$: $\vec{AP} = \langle (2\delta + 2) - 1, 0 - 1, (3\delta - 4) - (-1) \rangle$ $\vec{AP} = \langle 2\delta + 1, -1, 3\delta - 3 \rangle$

The direction vector of $L_2$ is $\vec{d_2} = \langle 2, 0, 3 \rangle$. Since $\vec{AP}$ is perpendicular to $L_2$, their dot product must be zero: $\vec{AP} \cdot \vec{d_2} = 0$ $(2\delta + 1)(2) + (-1)(0) + (3\delta - 3)(3) = 0$ $4\delta + 2 + 0 + 9\delta - 9 = 0$ $13\delta - 7 = 0$ $\delta = \frac{7}{13}$

Now, substitute $\delta = \frac{7}{13}$ back into the coordinates of $P$: $P = \left( 2\left(\frac{7}{13}\right) + 2, 0, 3\left(\frac{7}{13}\right) - 4 \right)$ $P = \left( \frac{14}{13} + \frac{26}{13}, 0, \frac{21}{13} - \frac{52}{13} \right)$ $P = \left( \frac{40}{13}, 0, -\frac{31}{13} \right)$

Step 4: Calculate $(\mathrm{PB})^2$

We need to find the squared distance between $P\left(\frac{40}{13}, 0, -\frac{31}{13}\right)$ and $B(4, 0, -1)$. $(\mathrm{PB})^2 = \left( \frac{40}{13} - 4 \right)^2 + (0 - 0)^2 + \left( -\frac{31}{13} - (-1) \right)^2$ $(\mathrm{PB})^2 = \left( \frac{40 - 52}{13} \right)^2 + 0 + \left( \frac{-31 + 13}{13} \right)^2$ $(\mathrm{PB})^2 = \left( \frac{-12}{13} \right)^2 + \left( \frac{-18}{13} \right)^2$ $(\mathrm{PB})^2 = \frac{144}{169} + \frac{324}{169}$ $(\mathrm{PB})^2 = \frac{144 + 324}{169} = \frac{468}{169}$ We can simplify this fraction: $468 = 13 \times 36$, and $169 = 13 \times 13$. $(\mathrm{PB})^2 = \frac{36 \times 13}{13 \times 13} = \frac{36}{13}$

Step 5: Calculate $26 \alpha (\mathrm{PB})^2$

We have $\alpha = 3$ and $(\mathrm{PB})^2 = \frac{36}{13}$. Substitute these values into the expression: $26 \alpha (\mathrm{PB})^2 = 26 \times 3 \times \frac{36}{13}$ $= (2 \times 13) \times 3 \times \frac{36}{13}$ $= 2 \times 3 \times 36$ $= 6 \times 36$ $= 216$

3. Common Mistakes & Tips

• Distinct Parameters: Always use different parameters (e.g., $\lambda, \mu, \delta$) when working with multiple lines or multiple points on the same line to avoid confusion and incorrect equations.
• Zero in Denominator: Understand that a zero in the denominator of a symmetric line equation (e.g., $\frac{z+1}{0}$) means the corresponding coordinate is constant (e.g., $z+1=0 \implies z=-1$).
• Dot Product for Perpendicularity: Remember that the dot product of two perpendicular vectors is zero. This is a crucial tool for finding the foot of a perpendicular.
• Arithmetic Precision: Be careful with fractions and calculations, especially when simplifying expressions involving squared distances.

4. Summary

We first converted the given lines into their parametric forms. By equating the coordinates of the general points on $L_1$ and $L_2$, we found the intersection point $B(4,0,-1)$ and the value of $\alpha=3$. Next, to find the foot of the perpendicular $P$ from $A(1,1,-1)$ to $L_2$, we represented $P$ parametrically and used the condition that $\vec{AP}$ is perpendicular to the direction vector of $L_2$. This allowed us to determine the coordinates of $P\left(\frac{40}{13}, 0, -\frac{31}{13}\right)$. Finally, we calculated the squared distance $(\mathrm{PB})^2$ and substituted all values into the expression $26 \alpha (\mathrm{PB})^2$ to arrive at the final result.

5. Final Answer

The value of $26 \alpha (\mathrm{PB})^2$ is $216$.

The final answer is $\boxed{216}$.