Write the given lines in vector form

$l_1: \vec r = \vec a_1 + \lambda \vec b_1$ where $\vec a_1=(1,-11,-7),\qquad \vec b_1=(1,2,3)$

$l_2: \vec r = \vec a_2 + \mu \vec b_2$ where $\vec a_2=(-1,0,1),\qquad \vec b_2=(2,2,1)$

Let the required line $l$ pass through origin and be perpendicular to both $l_1$ and $l_2$

Since $P$ is the intersection of $l$ with $l_1$, point $P$ lies on $l_1$ and also on $l$. As $l$ passes through origin, the direction of $l$ is along $\overrightarrow{OP}$.

Therefore, if $P=(1+\lambda,-11+2\lambda,-7+3\lambda),$ then $\overrightarrow{OP}=P$ must be perpendicular to the direction of $l_1$, i.e. $P\cdot \vec b_1=0.$

So, $(1+\lambda)+2(-11+2\lambda)+3(-7+3\lambda)=0$ $1+\lambda-22+4\lambda-21+9\lambda=0$ $14\lambda-42=0$ $\lambda=3.$

Hence, $P=(1+3,-11+6,-7+9)=(4,-5,2).$

Use the fact that $l$ is also perpendicular to $l_2$

Since $Q$ is the foot of perpendicular from $P$ to $l_2$, vector $\overrightarrow{PQ}$ is perpendicular to $l_2$. But line $l$ is perpendicular to $l_2$, and passes through $P$; hence the perpendicular from $P$ to $l_2$ is exactly along $l$.

Also, $l$ passes through origin, so origin lies on this perpendicular. Therefore the foot of the perpendicular from $P$ to $l_2$ must be the origin itself if origin lies on $l_2$.

Check whether origin lies on $l_2$: $(-1,0,1)+\mu(2,2,1)=(0,0,0).$ From coordinates: $-1+2\mu=0 \Rightarrow \mu=\frac12,$ $0+2\mu=0 \Rightarrow \mu=0,$ contradiction. So origin is not on $l_2$.

Hence we find $Q$ directly using projection.

Find foot of perpendicular from $P=(4,-5,2)$ on $l_2$

A general point on $l_2$ is $Q=(-1+2\mu,\ 2\mu,\ 1+\mu).$

Since $PQ \perp l_2$, we must have $(Q-P)\cdot (2,2,1)=0.$

Now $Q-P=(-1+2\mu-4,\ 2\mu+5,\ 1+\mu-2)=(2\mu-5,\ 2\mu+5,\ \mu-1).$

So, $(2\mu-5,2\mu+5,\mu-1)\cdot(2,2,1)=0$ $2(2\mu-5)+2(2\mu+5)+(\mu-1)=0$ $4\mu-10+4\mu+10+\mu-1=0$ $9\mu-1=0$ $\mu=\frac19.$

Therefore,

Compute the required value

$\alpha+\beta+\gamma=-\frac79+\frac29+\frac{10}{9}=\frac{5}{9}.$

Hence, $9(\alpha+\beta+\gamma)=9\cdot \frac59=5.$

Final Answer

$\boxed{5}$

Comparison with stored answer

Stored correct answer = $5$.

Our derived answer also equals $5$, so they agree.