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 _________ .

The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$$, from the origin is __________.

Let $$\mathrm{P}(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{Q}(1,6,4)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$. Then $$2 \alpha+\beta+\gamma$$ is equal to ________

If the shortest distance between the lines $$\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$$ and $$\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$$ is $$\frac{44}{\sqrt{30}}$$, then the largest possible value of $$|\lambda|$$ is equal to _________.