**24.** Let \[ L_1 : \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0} \quad \text{and} \quad L_2 : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha}, \ \alpha \in \mathbb{R} \] be two lines which intersect at the point \( B \). If \( P \) is the foot of the perpendicular from the point \( A(1, 1, -1) \) on \( L_2 \), then the value of \( 26\alpha (PB)^2 \) is _____.

Key ideas you must know

1. Symmetric form of a line in 3-D
   A line is often written as
   $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
   Here $(x_0,y_0,z_0)$ is any point on the line, and $(a,b,c)$ is its direction vector.

2. Intersection of two lines
   If two lines really meet, there will be parameter values (say $t$ on the first line and $s$ on the second) that give exactly the same $(x,y,z)$. Solve the coordinate equations to get both the common point B and maybe some unknown constant like α.

3. Foot of the perpendicular from a point to a line
   A point $P$ on line $L$ is the foot of the perpendicular from $A$ iff the vector $\overrightarrow{AP}$ is orthogonal (dot-product zero) to the line’s direction vector.

4. Distance between two points
   For $P(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$
   $PB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Typical logical chain a student would follow

• Identify direction vectors: $\mathbf{d_1}=(3,-1,0)$ for L1 and $\mathbf{d_2}=(2,0,α)$ for L2.
• Set up parameters $t,\;s$ so that the two lines are written in vector form.
  • Solve $x,y,z$ equations to find both the intersection point $B$ and the unknown $α$.
• With $α$ now known, write a general point $P$ on L2 using a fresh parameter (say $\lambda$).
• Impose the perpendicularity condition $\overrightarrow{AP}\,\cdot\,\mathbf{d_2}=0$ to fix $\lambda$; this gives the exact coordinates of $P$.
• Compute $PB$, square it, multiply by $26α$, and simplify. Every step is just straight textbook 3-D coordinate geometry.

What’s happening here?

Imagine two long sticks in space, named L1 and L2. They touch each other at one common point B (so they really do meet). A little ant is standing at point A(1,1,−1). The ant wants to walk straight to stick L2 in the shortest possible way – that means along a line perpendicular (90°) to stick L2. The point where the ant lands is P (the foot of the perpendicular).

The question finally asks for a special number:

First we must find how far P is from B (the length PB).

Then we multiply the square of that length by 26 × α, where α is a small hidden number sitting in the equation of line L2.

Do the maths right, and you will get one nice integer.

Step-by-step solution

1. Write both lines in parametric (vector) form

   L_1 :\; & x = 1 + 3t, && y = 1 - t, && z = -1 + 0t\\ L_2 :\; & x = 2 + 2s, && y = 0 + 0s, && z = -4 + αs \end{aligned}$$

2. Find the intersection point $B$ and $α$

   At intersection, $(x,y,z)$ from both lines must be equal for some $t=something$ and $s=something$.

   From the $y$–coordinate:
   $1 - t = 0 \;\Rightarrow\; t = 1$

   From the $x$–coordinate:
   $1 + 3(1) = 2 + 2s \;\Rightarrow\; 4 = 2 + 2s \;\Rightarrow\; s = 1$

   From the $z$–coordinate:
   $-1 = -4 + α(1) \;\Rightarrow\; α = 3$

   Therefore $B(4,\,0,\,-1)$

3. Direction vector of $L_2$ with found $α$
   $\mathbf{d_2} = (2,\,0,\,3)$

4. General point $P$ on $L_2$
   Let parameter be $\lambda$: $P(\,2+2\lambda,\;0,\;-4+3\lambda\,)$

5. Perpendicularity condition ($\overrightarrow{AP}\perp L_2$)

   Vector $\overrightarrow{AP} = (2+2\lambda-1,\;0-1,\;-4+3\lambda+1) = (1+2\lambda,\,-1,\,-3+3\lambda)$

   Dot product with $\mathbf{d_2}$ must vanish:

   (1+2\lambda, -1, -3+3\lambda) \cdot (2,0,3) &= 2(1+2\lambda) + 0 + 3(-3+3\lambda)\\ &= 2 + 4\lambda - 9 + 9\lambda\\ &= 13\lambda - 7 = 0 \end{aligned}$$ $$\Rightarrow \lambda = \frac{7}{13}$$

6. Coordinates of $P$

   x_P &= 2 + 2\left(\frac{7}{13}\right) = \frac{40}{13}\\ y_P &= 0\\ z_P &= -4 + 3\left(\frac{7}{13}\right) = -\frac{31}{13} \end{aligned}$$

7. Vector $\overrightarrow{PB}$ and its square length

   = \left(\frac{12}{13},\;0,\;\frac{18}{13}\right)$$ $$\begin{aligned} (PB)^2 &= \left(\frac{12}{13}\right)^2 + 0^2 + \left(\frac{18}{13}\right)^2\\ &= \frac{144 + 324}{169} = \frac{468}{169} \end{aligned}$$

8. Compute the required expression

   $26\alpha (PB)^2 = 26 \times 3 \times \frac{468}{169}$

   Note that $26/169 = 2/13$: $26\alpha (PB)^2 = 3 \times 468 \times \frac{2}{13} = \frac{2808}{13} = 216$

Final Answer: 216