Let a line $$L$$ pass through the origin and be perpendicular to the lines
$$L_1: \vec{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$$, $$\lambda \in \mathbb{R}$$ and
$$L_2: \vec{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$$, $$\mu \in \mathbb{R}$$. If $$P$$ is the point of intersection of $$L$$ and $$L_1$$, and $$Q(\alpha, \beta, \gamma)$$ is the foot of perpendicular from $$P$$ on $$L_2$$, then $$9\alpha + \beta + \gamma$$ is equal to _______

Line $$L$$ passes through the origin and is perpendicular to $$L_1$$ (direction $$\hat{i} + 2\hat{j} + 3\hat{k}$$) and $$L_2$$ (direction $$2\hat{i} + 2\hat{j} + \hat{k}$$). $$P$$ is the intersection of $$L$$ and $$L_1$$, and $$Q(\alpha,\beta,\gamma)$$ is the foot of the perpendicular from $$P$$ to $$L_2$$. We wish to find $$9\alpha + \beta + \gamma$$.

Since $$L$$ is perpendicular to both $$L_1$$ and $$L_2$$, its direction is: $$\vec{d}_L = (1,2,3)\times(2,2,1) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1\end{vmatrix}$$ which simplifies to $$\hat{i}(2-6) - \hat{j}(1-6) + \hat{k}(2-4) = -4\hat{i} + 5\hat{j} - 2\hat{k}.$$ Thus $$L: \vec{r} = t(-4\hat{i} + 5\hat{j} - 2\hat{k}),\; t\in\mathbb{R}.$$

For the intersection with $$L_1$$, which is given by $$(1+\lambda,\,-11+2\lambda,\,-7+3\lambda),$$ we set $$(-4t,\,5t,\,-2t) = (1+\lambda,\,-11+2\lambda,\,-7+3\lambda).$$ This yields the system $$-4t = 1+\lambda,\quad 5t = -11+2\lambda,\quad -2t = -7+3\lambda.$$ From the first equation, $$\lambda=-4t-1$$. Substitution into the second gives $$5t = -11 + 2(-4t-1) = -13 - 8t\implies 13t = -13\implies t=-1,$$ hence $$\lambda = 3$$. Verification in the third equation, $$-2(-1)=2\quad\text{and}\quad -7+9=2,$$ confirms consistency. Therefore $$P = (4,\,-5,\,2).$$

To find the foot of the perpendicular from $$P$$ to $$L_2$$, note that $$L_2: (-1+2\mu,\,2\mu,\,1+\mu)$$ and $$\overrightarrow{PQ} = \bigl(-1+2\mu-4,\;2\mu+5,\;1+\mu-2\bigr) = (-5+2\mu,\;5+2\mu,\;\mu-1).$$ Requiring $$\overrightarrow{PQ}\cdot\vec{d}_2=0$$ with $$\vec{d}_2=(2,2,1)$$ gives $$2(-5+2\mu) + 2(5+2\mu) + 1(\mu-1) = 0 \implies -10+4\mu+10+4\mu+\mu-1 = 0 \implies 9\mu - 1 = 0 \implies \mu = \frac{1}{9}.$$

Substituting $$\mu=\tfrac{1}{9}$$ into the parametric form of $$L_2$$ yields $$\alpha = -1 + \frac{2}{9} = -\frac{7}{9},\quad \beta = \frac{2}{9},\quad \gamma = 1 + \frac{1}{9} = \frac{10}{9}.$$

Answer: $$9\alpha + \beta + \gamma = 9\Bigl(-\frac{7}{9}\Bigr) + \frac{2}{9} + \frac{10}{9} = -7 + \frac{12}{9} = -7 + \frac{4}{3} = -\frac{17}{3}.$$