<s>[INST] If the distance between two lines is $ \frac{1}{\sqrt{3}} $, then $ \alpha $ is... 
If the shortest distance between the lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$ and
$x+y+z+1=0 = 2x-y+z+3 = 0$ is $\displaystyle \frac{1}{\sqrt{3}}\;,$ Then $\alpha = $

$\bf{My\; Solution:: }$ Here $x+y+z+1=0$ and $2x-y+z+3=0$ represent $2$ plane and
we know that locus of  point of Intersection of $2$ plane is also a Line.
Know Here we have to calculate the equation of Common line of these $2$ planes.
For Calculation of equation of line....
Add these $2$ equations... $\displaystyle 3x+2z+4=0\Rightarrow z= - \frac{\left(4+3x\right)}{2}$ and Put into $x+y+z+1=0$
$\displaystyle 2x+2y+2z+2=0\Rightarrow 2x+2y-4-3x+2=0\Rightarrow -x+2y=2$
So we get $\displaystyle x=2y-2\Rightarrow \frac{x-0}{1} = \frac{y-1}{\frac{1}{2}}$ and above we get $\displaystyle \frac{x-0}{1}=\frac{z+2}{-\frac{3}{2}}.$
So We get equation of line is $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Now we have to Calculate Distance b/w two lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$
and $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Is my equation of line is Right or not, If not then how can i calculate common
equation of line and value of $\alpha.$
Help me , Thanks
 [/INST] I would go about doing it like this: put your lines into vector notation, in this case, $$(x_1,y_1,z_1)=(0,1,0)+t_1(2,1,0)$$ $$(x_2,y_2,z_2)=(1,-1,0)+t_2(\alpha,-1,1)$$ Then solve the messy system of equations yielded by$$\frac{\partial}{\partial t_1 }||(x_1,y_1,z_2)-(x_2,y_2,z_2)||^2=0$$ $$\frac{\partial}{\partial t_2 }||(x_1,y_1,z_2)-(x_2,y_2,z_2)||^2=0$$ $$||(x_1,y_1,z_1)-(x_2,y_2,z_2)||^2=\frac{1}{3}$$
for $\alpha$.
Edit: Alternatively, you could use the fact that if you project the vector connecting a point on one line and the other onto the vector normal to both, you'll get the distance between them. That is, $$((0,1,0)-(1,-1,0))\cdot\left(\frac{(2,1,0)\times(\alpha,-1,1)}{||(2,1,0)\times(\alpha,-1,1)||}\right)=\alpha$$ where $||\textbf{v}||$ is the euclidean norm of $\textbf{v}$.
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