Let a line $l$ pass through the origin and be perpendicular to the lines

$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and

$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.

If $\mathrm{P}$ is the point of intersection of $l$ and $l_{1}$, and $\mathrm{Q}(\propto, \beta, \gamma)$ is the foot of perpendicular from P on $l_{2}$, then $9(\alpha+\beta+\gamma)$ is equal to _____________.

We have,<br/> <p>$$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$$ and</p> <p>$$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$$.</p> Let direction ratio of line $l$ be $a, b$ and $c$ Equation of line $l$ <br/><br/>$$ \begin{aligned} \vec{r} & =(0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\\\ & =\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \end{aligned} $$ <br/><br/>As, line $l$ is perpendicular to $l_1$ and $l_2$, <br/><br/>$$ a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{array}\right|=-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} $$ <br/><br/>$\therefore$ Equation of line $l: \vec{r}=\delta(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ <br/><br/>As, $P$ is the intersecting point of $l$ and $l_1$ <br/><br/>$-4 \delta=1+\lambda, 5 \delta=-11+2 \lambda,-2 \delta=-7+3 \lambda$ <br/><br/>After solving the above three equation, we get <br/><br/>$\delta=-1 \text { and } \lambda=3$ <br/><br/>$\therefore$ Co-ordinate of point $P$ is $(4,-5,2)$. <br/><br/>$Q$ is a point on line $l_2$ <br/><br/>Let co-ordinate of $Q$ be $(-1+2 \mu, 2 \mu, 1+\mu)$ <br/><br/>$$ \begin{gathered} \overrightarrow {PQ}=(-5+2 \mu) \hat{\mathbf{i}}+(2 \mu+5) \hat{\mathbf{j}}+(\mu-1) \hat{\mathbf{k}} \\\\ \overrightarrow {PQ} \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \left[\because \overrightarrow {PQ} \perp l_2\right]\\\\ 2(-5+2 \mu)+2(2 \mu+5)+\mu-1=0 \\\\ 9 \mu-1=0 \Rightarrow \mu=\frac{1}{9} \end{gathered} $$ <br/><br/>$$ \begin{aligned} & \therefore \alpha=-1+\frac{2}{9}=\frac{-7}{9}, \beta=2 \times \frac{1}{9}=\frac{2}{9}, r=1+\frac{1}{9}=\frac{10}{9} \\\\ & \text { Hence, } 9(\alpha+\beta+\gamma)=9\left(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9}\right)=5 \end{aligned} $$