Let $$L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$$ and $$L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}$$, $$\alpha \in R$$, 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 $$26\alpha(PB)^{2}$$ is ______

We have $$L_1: \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$$ and $$L_2: \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha}$$.

Find the intersection point $$B$$.

Parametric forms: $$L_1: (1+3t,\; 1-t,\; -1)$$ and $$L_2: (2+2s,\; 0,\; -4+\alpha s)$$.

Setting coordinates equal:

From $$y$$: $$1 - t = 0 \implies t = 1$$

From $$x$$: $$1 + 3(1) = 2 + 2s \implies s = 1$$

From $$z$$: $$-1 = -4 + \alpha(1) \implies \alpha = 3$$

So $$B = (4, 0, -1)$$ and $$\alpha = 3$$.

Find foot of perpendicular $$P$$ from $$A(1, 1, -1)$$ on $$L_2$$.

A general point on $$L_2$$: $$(2 + 2\delta,\; 0,\; -4 + 3\delta)$$.

$$\vec{AP} = (1 + 2\delta,\; -1,\; -3 + 3\delta)$$

Direction of $$L_2$$: $$(2, 0, 3)$$.

For perpendicularity: $$\vec{AP} \cdot (2, 0, 3) = 0$$

$$2(1 + 2\delta) + 0 + 3(-3 + 3\delta) = 0$$

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

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

Compute $$(PB)^2$$.

$$\vec{PB} = \left(4 - \frac{40}{13},\; 0,\; -1 + \frac{31}{13}\right) = \left(\frac{12}{13},\; 0,\; \frac{18}{13}\right)$$

$$(PB)^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}$$

Calculate $$26\alpha(PB)^2$$.

$$26 \times 3 \times \frac{468}{169} = \frac{36504}{169} = 216$$

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