Let a line $l$  pass through the origin and be perpendicular to the lines ${\mathrm{l}}_1:\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-11\hat{\mathrm{j}}-7\hat{\mathrm{k}})+\mathrm{\lambda }(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}),\mathrm{\lambda }\in \mathrm{ℝ}$ and ${\mathrm{l}}_2:\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+\hat{\mathrm{k}})+\mathrm{\mu }(2\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}),\mathrm{\mu }\in \mathrm{ℝ}$
If P is the point of intersection of l and l₁ , and $\mathrm{Q}(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ is the foot of the perpendicular from P on l₂, then $9(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })$ is equal to _______

$\mathcal{l}\equiv 0$ passing through $\mathrm{O}(0,0,0)$ and perpendicular to the lines
$\begin{array}{l}{\mathcal{l}}_1:\overline{\mathrm{r}}=(1,-11,-7)+\mathrm{\lambda }(1,2,3),\mathrm{\lambda }\in \mathrm{ℝ}\\ {\mathcal{l}}_2:\overline{\mathrm{r}}=(-1,0,1)+\mathrm{\mu }(2,2,1),\mathrm{\mu }\in \mathrm{ℝ}\end{array} \mathcal{l}:\frac{\mathrm{x}-0}{\mathrm{a}}=\frac{\mathrm{y}-0}{\mathrm{b}}=\frac{\mathrm{z}-0}{\mathrm{c}},$
$$\begin{gathered} {\mathcal{l}}_1:\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+11}{2}=\frac{\mathrm{z}+7}{3}=\mathrm{\lambda } {\mathcal{l}}_2:\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-0}{2}=\frac{\mathrm{z}-1}{1}=\mathrm{\mu } , \\ \Rightarrow \mathrm{a}+2\mathrm{b}+3\mathrm{c}=0 \end{gathered}$$

$2\mathrm{a}+2\mathrm{b}+\mathrm{c}=0$

$\frac{\mathrm{a}}{2-6}=\frac{\mathrm{b}}{6-1}=\frac{\mathrm{c}}{2-4}\Rightarrow \frac{\mathrm{a}}{-4}=\frac{\mathrm{b}}{5}=\frac{\mathrm{c}}{-2}$
$\therefore \mathcal{l}=\frac{\mathrm{x}}{4}=\frac{\mathrm{y}}{-5}=\frac{\mathrm{z}}{2}=\mathrm{t} \mathrm{P}.\mathrm{I}\mathrm{of}\mathcal{l}=0\&{\mathcal{l}}_1=0$
$$\begin{gathered} (4\mathrm{t},-5\mathrm{t},2\mathrm{t})=(\mathrm{\lambda }+1,2\mathrm{\lambda }-11,3\mathrm{\lambda }-7)\Rightarrow 4\mathrm{t}=\mathrm{\lambda }+1,-5\mathrm{t}=2\mathrm{\lambda }-11,2\mathrm{t}=3\mathrm{\lambda }-7 \\ \Rightarrow 2\mathrm{t}=\frac{\mathrm{\lambda }+1}{2} 3\mathrm{\lambda }-7=\frac{\mathrm{\lambda }+1}{2} \begin{array}{l}\Rightarrow 6\mathrm{\lambda }-14=\mathrm{\lambda }+1\\ \Rightarrow 5\mathrm{\lambda }=15\\ \Rightarrow \boxed{\mathrm{\lambda }=3}\boxed{\mathrm{t}=1}\end{array} \end{gathered}$$
$\therefore \mathrm{P}.\mathrm{I}\mathrm{\hspace{0.17em}}\mathrm{of}\mathrm{\hspace{0.17em}}\mathcal{l}=0,{\mathcal{l}}_1=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(4,-5,2)$

$\mathrm{Q}(2\mathrm{\mu }-1,2\mathrm{\mu },\mathrm{\mu }+1)$ be a point on line ${\mathcal{l}}_2=0$
Direction ratios of PQ: $(2\mathrm{\mu }-1-4,2\mathrm{\mu }+5,\mathrm{\mu }+1-2)$
$=(2\mathrm{\mu }-5,2\mathrm{\mu }+5,\mathrm{\mu }-1)$ 

Direction ratios of  line $(2,2,1)$PQ perpendicular to ${\mathcal{l}}_2=0$
$4\mathrm{\mu }-10+4\mathrm{\mu }+10+\mathrm{\mu }-1=0$

$\mathrm{\mu }=\frac{1}{9}\Rightarrow \mathrm{Q}(\frac{2}{9}-1,\frac{2}{9},\frac{1}{9}+1)=(-\frac{7}{9},\frac{2}{9},\frac{10}{9})$
$\therefore 9(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })=5$