Solution

Shortest distance between the lines $$\displaystyle \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}, \,\,\, (\alpha \neq -1)$$ and $$x + y + z + 1 = 0 = 2x - y + z + 3$$ is $$\displaystyle \frac{1}{\sqrt{3}}$$ Line of intersection of planes $$x + y + z + 1 = 0 = 2x - y + z + 3$$ is Eliminating $$y$$ gives $$3x+2z+4=0$$ $$\Rightarrow \displaystyle x=\frac{-2z-4}{3}$$ ----(1) Substituting above $$x$$ in $$x+y+z+1=0$$ gives $$3y+z-1=0$$ $$\Rightarrow z=-3y+1$$ ----(2) From (1) and (2), line equation is $$\displaystyle\frac{3x+4}{-2}=-3y+1=z$$ $$\Rightarrow \displaystyle\frac{x-(-4/3)}{-2/3}=\frac{y-(1/3)}{-1/3}=\frac{z}{1}$$ Given line is $$\displaystyle \frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}$$ Shortest distance between above two skew lines is $$\displaystyle\frac{(\vec b-\vec a)\cdot(\vec c\times \vec d)}{|\vec c\times\vec d|}$$ where one line is passing through $$\vec a$$ and parallel to $$\vec c$$ and other one passes through $$\vec b$$ and parallel to $$\vec d$$ $$\vec a =(1,-1,0)$$ $$\vec b=\left (-\dfrac {4}{3},\dfrac {1}{3},0\right)$$ $$\vec c=(\alpha,-1,1)$$ $$\vec d=\left (-\dfrac {2}{3},-\dfrac {1}{3},1\right)$$ $$\therefore \displaystyle\frac{(\vec b-\vec a)\cdot(\vec c\times \vec d)}{|\vec c\times\vec d|}=\displaystyle \left| \frac { \begin{vmatrix} 7/3 & -4/3 & 0 \\ \alpha & -1 & 1 \\ -2/3 & -1/3 & 1 \end{vmatrix} }{ \begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ \alpha & -1 & 1 \\ -2/3 & -1/3 & 1 \end{vmatrix} } \right| =\frac { 1 }{ \sqrt { 3 } } $$ Expanding determinants and simplifying gives $$48\alpha^2-48\alpha+12=10\alpha^2+16\alpha+12$$ $$\Rightarrow 38\alpha^2=64\alpha$$ $$ \displaystyle \Rightarrow \alpha=\frac{64}{38}=\frac{32}{19}$$ Hence, option A.