A general point on \(L_1\) is \((1+3t,\ 1-t,\ -1)\) and on \(L_2\) is \((2+2s,\ 0,\ -4+\alpha s)\). For intersection, equating components: \(1-t=0\Rightarrow t=1\), so the point is \((4,0,-1)\). From \(L_2\): \(2+2s=4\Rightarrow s=1\) and \(-4+\alpha=-1\Rightarrow\alpha=3\). So \(B=(4,0,-1)\) and \(L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{3}\). Foot of perpendicular from \(A(1,1,-1)\) to \(L_2\): direction of \(L_2\) is \(\vec{d}=(2,0,3)\). Point on \(L_2\): \((2+2s,0,-4+3s)\). Vector from this point to \(A\): \((1-(2+2s),\ 1-0,\ -1-(-4+3s))=(-1-2s,1,3-3s)\). Dot with \(\vec{d}=0\): \(2(-1-2s)+0+3(3-3s)=0\Rightarrow-2-4s+9-9s=0\Rightarrow13s=7\Rightarrow s=\frac{7}{13}\). So \(P=\left(2+\frac{14}{13},0,-4+\frac{21}{13}\right)=\left(\frac{40}{13},0,-\frac{31}{13}\right)\). \(PB^2=\left(4-\frac{40}{13}\right)^2+(0)^2+\left(-1+\frac{31}{13}\right)^2=\left(\frac{12}{13}\right)^2+\left(\frac{18}{13}\right)^2=\frac{144+324}{169}=\frac{468}{169}=\frac{36}{13}\). Therefore \(26\alpha\cdot(PB)^2=26\times3\times\frac{36}{13}=6\times36=216\).