Let a line $L$ pass through the origin and be perpendicular to the lines $L_1: \vec{r} = \hat{i} - 11\hat{j}…

Given,

A line $L$ pass through the origin and be perpendicular to the lines $L_1:\overrightarrow{r}=\left(\hat{i}-11\hat{j}-7\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right), \lambda \in \mathrm{ℝ}$ and
$L_2:\overrightarrow{r}=\left(-\hat{i}+\hat{k}\right)+\mu \left(2\hat{i}+2\hat{j}+\hat{k}\right), \mu \in \mathrm{ℝ}$

So, direction ratio of line will be,
$ \begin{aligned} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{vmatrix} \end{aligned}$

$=\hat{i}\left(-4\right)-\hat{j}\left(-5\right)+\hat{k}\left(-2\right)$

$=-4\hat{i}+5\hat{j}-2\hat{k}$

Hence, the equation of line $L$ which passes through origin will be,

$L:\overrightarrow{r}=\sigma (-4\hat{i}+5\hat{j}-2\hat{k})$

Now finding the intersection point $P$ we get,

$1+\lambda =-4\sigma ....\left(1\right)$

$-11+2\lambda =5\sigma ....\left(2\right)$

$-7+3\lambda =-2\sigma ....\left(3\right)$

So, from equation $\left(1\right) \& \left(3\right)$ we get, $1+\lambda =-14+6\lambda \Rightarrow \lambda =3, \sigma =-1$

Hence, point $P\equiv \left(4,-5,2\right)$

Now finding the point $Q$ which is foot of perpendicular from point $P\left(4,-5,2\right)$ on line $L_2:\hspace{0.17em}\left(-\hat{i}+\hat{k}\right)+\mu \left(2\hat{i}+2\hat{j}+\hat{k}\right)$ we get,

Now any point on line $L_2$ will be $Q\left(\left(2\mu -1\right)\hat{i}+2\mu \hat{j}+\left(\mu +1\right)\hat{k}\right)$

Now direction ratio of $PQ$ will be $\left(-5+2\mu \right)\hat{i}+\left(2\mu +5\right)\hat{j}+\left(\mu -1\right)\hat{k}=0$

Now using the perpendicular condition we get,

$2(-5+2\mu )+2(2\mu +5)+1(\mu -1)=0$

$\Rightarrow \mu =\frac{1}{9}$

Hence, $\alpha +\beta +\gamma =2\mu -1+2\mu +\mu +1=5\mu =\frac{5}{9}$

$\Rightarrow 9\left(\alpha +\beta +\gamma \right)=5$