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
 A: 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 $\textbf{v}$.
A: I like to use Plücker coordinates for lines


*

*Line along $(\alpha,-1,1)$,  through Point $(1,-1,0)$


$$ L_1 = \left\{ \begin{array}{c} \vec{e} \\ \vec{r}\times\vec{e} \end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix} \alpha\\-1\\1\end{pmatrix} \\
\begin{pmatrix} 1\\-1\\0\end{pmatrix} \times \begin{pmatrix} \alpha\\-1\\1\end{pmatrix}
\end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix} \alpha\\-1\\1\end{pmatrix} \\ \begin{pmatrix} -1\\-1\\ \alpha-1 \end{pmatrix} \end{array} \right\} =\left\{ \begin{array}{c} \vec{\ell}_1 \\ \vec{m}_1 \end{array} \right\} $$


*

*Line where Plane $\vec{n_A}=(-1,-1,-1)$, $d_A=1$ and Plane $\vec{n_B}=(-2,1,-1)$, $d_B=3$ meet
$$ L_2 = \left\{ \begin{array}{c} \vec{n}_A \times \vec{n}_B \\ d_B \vec{n}_A-d_A \vec{n}_A \end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \times \begin{pmatrix}-2\\1\\-1\end{pmatrix} \\ 3 \begin{pmatrix}-1\\-1\\-1\end{pmatrix}-1 \begin{pmatrix}-2\\1\\-1\end{pmatrix}  \end{array} \right\}=\left\{ \begin{array}{c} \begin{pmatrix}2\\1\\-3\end{pmatrix} \\ \begin{pmatrix}-1\\-4\\-2\end{pmatrix}   \end{array}\right\} =\left\{ \begin{array}{c} \vec{\ell}_2 \\  \vec{m}_2 \end{array}\right\} $$

*Minimum distance between lines $L_1$ and $L_2$
$$ \rho = \frac{\vec{m}_1 \cdot \vec{\ell}_2 + \vec{m}_2 \cdot \vec{\ell}_1} {\| \vec{\ell}_1 \times \vec{\ell}_2 \| } = \frac{\begin{pmatrix} -1 \\ -1 \\ \alpha-1 \end{pmatrix}\cdot \begin{pmatrix} 2 \\ 1 \\-3 \end{pmatrix}+\begin{pmatrix}-1\\-4\\-2 \end{pmatrix}\cdot\begin{pmatrix}\alpha\\-1\\1 \end{pmatrix}}{\| \begin{pmatrix} \alpha\\-1\\1\end{pmatrix}\times \begin{pmatrix}2\\1\\-3 \end{pmatrix} \|} = \frac{2-4 \alpha}{\sqrt{10 \alpha^2+16 \alpha+12}} $$
We know that $\rho=1/\sqrt{3}$ and thus
$$\left. \frac{1}{\sqrt{3}} = \frac{2-4 \alpha}{\sqrt{10 \alpha^2+16 \alpha+12}}  \right\} \boxed{ \alpha=0 \;\&\; \alpha=\frac{32}{19}}$$
From the two solutions we pick $\alpha=0$ (simplest solution) to get an additional level of detail:


*

*Line normal to both lines $L_1$ and $L_2$ through the shortest distance (for reference)


$$ L_3 =\left\{ \begin{array}{c} \vec{\ell}_1 \times \vec{\ell}_2 \\ \vec{m}_1 \times \vec{\ell}_2 - \vec{m}_2 \times \vec{\ell}_1 \end{array} \right\} =
\left\{ \begin{array}{c} \begin{pmatrix}2\\2\\ 2\end{pmatrix} \\ \begin{pmatrix} 10 \\ -6 \\ 0 \end{pmatrix}   \end{array}\right\}  =\left\{ \begin{array}{c} \vec{\ell}_3 \\ \vec{m}_3 \end{array} \right\}$$


*

*The direction $\vec{e}_3$ and position $\vec{r}_3$ of the cross line is


$$ \vec{e}_3 = \frac{ \vec{\ell}_3 }{ \| \vec{\ell}_3 \| } = \begin{pmatrix}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} $$
$$ \vec{r}_3 = \frac{ \vec{\ell}_3 \times \vec{m}_3 }{\| \vec{\ell}_3 \|^2} = \begin{pmatrix}1\\ \frac{5}{3} \\ -\frac{8}{3} \end{pmatrix}$$


*

*Confirm results with GeoGebra 3D


Distance AB is $1/\sqrt{3} = 0.5774$

