Let  $L_1$  and $L_2$  be the following straight lines.
$L_1:\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and  $L_2:\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$. Suppose the straight line $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$  lies in the plane containing $L_1$  and $L_2$ and passes through the point of intersection of $L_1$ and  $L_2$. If the line L bisects the acute angle between the lines $L_1$  and $L_2$ , then the value of  $(2\alpha +\gamma +l+m)$  is_______

Vector equation of the given straight lines are
$r=(\widehat{i}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+3\widehat{k})$   and  $r=(\widehat{i}+\widehat{k})+v(-3\widehat{i}-\widehat{j}+\widehat{k})$
$\because (\widehat{i}-\widehat{j}+3\widehat{k}).(-3\widehat{i}-\widehat{j}+\widehat{k})=-3+1+3=1$ is positive,
$\therefore$  Angle between supporting line vectors of lines $L_1$ and $L_2$  is acute, and point of intersection of given lies $L_1$  and $L_2$  is (1, 0, 1).
vector along the acute angle bisector of vectors
$(\widehat{i}-\widehat{j}+3\widehat{k})$  and $(-3\widehat{i}-\widehat{j}+\widehat{k})$  is $(\widehat{i}-\widehat{j}+2\widehat{k})$ or  $(\widehat{i}+\widehat{j}-2\widehat{k})$.
It is given that line $L:\frac{x-\alpha }{l}=\frac{y-1}{m}=\frac{z-\gamma }{-2}$  is the bisector of the acute angle between the lines $L_1$ and $L_2$ , so $l=1$ and  $m=1$ and  $\frac{1-\alpha }{1}=\frac{0-1}{1}=\frac{1-\gamma }{-2}$
$\Rightarrow \alpha =2,\gamma =-1 \therefore \alpha -\gamma =3,l+m=2$