1. Key Concepts and Formulas

• Direction Vector of a Line Perpendicular to Two Lines: If a line $l$ is perpendicular to two lines $l_1$ and $l_2$, its direction vector $\vec{d_l}$ is parallel to the cross product of the direction vectors of $l_1$ and $l_2$, i.e., $\vec{d_l} \propto (\vec{d_1} \times \vec{d_2})$.
• Point of Intersection of Two Lines: A point of intersection satisfies the parametric equations of both lines. Equating the coordinates provides a system of linear equations to solve for the parameters.
• Foot of Perpendicular from a Point to a Line: If $Q$ is the foot of the perpendicular from point $P$ to line $L$, then the vector $\vec{PQ}$ is perpendicular to the direction vector of line $L$. This implies their dot product is zero: $\vec{PQ} \cdot \vec{d_L} = 0$.

2. Step-by-Step Solution

Step 1: Identify Given Information and Direction Vectors We are given two lines, $l_1$ and $l_2$, in vector form:

• Line $l_1$: $\vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k})$ This tells us that $l_1$ passes through the point $A_1(1, -11, -7)$ and has a direction vector $\vec{d_1} = \hat{i}+2 \hat{\jmath}+3 \hat{k}$.
• Line $l_2$: $\vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}})$ This tells us that $l_2$ passes through the point $A_2(-1, 0, 1)$ and has a direction vector $\vec{d_2} = 2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}$. Line $l$ passes through the origin $O(0,0,0)$ and is perpendicular to both $l_1$ and $l_2$.

Step 2: Find the Direction Vector and Equation of Line $l$

• Why: Line $l$ is perpendicular to both $l_1$ and $l_2$, so its direction vector must be perpendicular to $\vec{d_1}$ and $\vec{d_2}$. The cross product of $\vec{d_1}$ and $\vec{d_2}$ will give us such a vector.
• Calculation: Let $\vec{d_l}$ be the direction vector of line $l$. $\vec{d_l} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{\mathrm{k}} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{vmatrix}$ $\vec{d_l} = \hat{\imath}((2)(1) - (3)(2)) - \hat{\jmath}((1)(1) - (3)(2)) + \hat{\mathrm{k}}((1)(2) - (2)(2))$ $\vec{d_l} = \hat{\imath}(2 - 6) - \hat{\jmath}(1 - 6) + \hat{\mathrm{k}}(2 - 4)$ $\vec{d_l} = -4\hat{\imath} + 5\hat{\jmath} - 2\hat{\mathrm{k}}$
• Reasoning: Line $l$ passes through the origin $O(0,0,0)$ and has direction vector $\vec{d_l}$. Its vector equation is $\vec{r} = \delta \vec{d_l}$. $\vec{r} = \delta(-4\hat{\imath} + 5\hat{\jmath} - 2\hat{\mathrm{k}}), \delta \in \mathbb{R}$ Any point on line $l$ can be represented as $(-4\delta, 5\delta, -2\delta)$.

Step 3: Find the Point of Intersection $P$ of $l$ and $l_1$

• Why: The point $P$ lies on both lines $l$ and $l_1$. We equate their parametric forms to find the values of the parameters $\delta$ and $\lambda$.
• Calculation: A general point on $l$ is $P(-4\delta, 5\delta, -2\delta)$. A general point on $l_1$ is $P(1+\lambda, -11+2\lambda, -7+3\lambda)$. Equating the coordinates:
  1. $-4\delta = 1+\lambda$
  2. $5\delta = -11+2\lambda$
  3. $-2\delta = -7+3\lambda$ From equation (1), $\lambda = -4\delta - 1$. Substitute this into equation (2): $5\delta = -11 + 2(-4\delta - 1)$ $5\delta = -11 - 8\delta - 2$ $13\delta = -13 \implies \delta = -1$. Substitute $\delta = -1$ back into the expression for $\lambda$: $\lambda = -4(-1) - 1 = 4 - 1 = 3$.
• Reasoning: We must verify these parameter values with the third equation. For equation (3): $-2(-1) = 2$ and $-7+3(3) = -7+9 = 2$. The values are consistent. Now, substitute $\delta = -1$ into the parametric form of $P$ on line $l$: $P = (-4(-1), 5(-1), -2(-1)) = (4, -5, 2)$.

Step 4: Find the Foot of Perpendicular $Q$ from $P$ on $l_2$

• Why: The foot of the perpendicular $Q$ is a point on $l_2$ such that the vector $\vec{PQ}$ is perpendicular to the direction vector of $l_2$. We use the dot product condition $\vec{PQ} \cdot \vec{d_2} = 0$.
• Calculation: Let $Q(\alpha, \beta, \gamma)$ be a general point on $l_2$: $Q(-1+2\mu, 2\mu, 1+\mu)$. The point $P$ is $(4, -5, 2)$. The vector $\vec{PQ}$ is $Q-P$: $\vec{PQ} = ((-1+2\mu) - 4)\hat{\imath} + (2\mu - (-5))\hat{\jmath} + ((1+\mu) - 2)\hat{\mathrm{k}}$ $\vec{PQ} = (2\mu - 5)\hat{\imath} + (2\mu + 5)\hat{\jmath} + (\mu - 1)\hat{\mathrm{k}}$ The direction vector of $l_2$ is $\vec{d_2} = 2\hat{\imath} + 2\hat{\jmath} + \hat{\mathrm{k}}$. Apply the perpendicularity condition $\vec{PQ} \cdot \vec{d_2} = 0$: $(2\mu - 5)(2) + (2\mu + 5)(2) + (\mu - 1)(1) = 0$ $4\mu - 10 + 4\mu + 10 + \mu - 1 = 0$ $9\mu - 1 = 0 \implies \mu = \frac{1}{9}$
• Reasoning: Substitute $\mu = \frac{1}{9}$ into the parametric form of $Q$ to find its coordinates: $\alpha = -1 + 2\left(\frac{1}{9}\right) = -1 + \frac{2}{9} = -\frac{7}{9}$ $\beta = 2\left(\frac{1}{9}\right) = \frac{2}{9}$ $\gamma = 1 + \frac{1}{9} = \frac{10}{9}$ So, $Q = (-\frac{7}{9}, \frac{2}{9}, \frac{10}{9})$.

Step 5: Calculate $9(\alpha+\beta+\gamma)$

• Why: We need to find the sum of the coordinates of $Q$ and multiply by 9, as requested by the problem.
• Calculation: $\alpha+\beta+\gamma = -\frac{7}{9} + \frac{2}{9} + \frac{10}{9} = \frac{-7+2+10}{9} = \frac{5}{9}$ Finally, calculate $9(\alpha+\beta+\gamma)$: $9(\alpha+\beta+\gamma) = 9 \left(\frac{5}{9}\right) = 5$

3. Common Mistakes & Tips

• Cross Product Calculation: Be careful with signs and determinant expansion. A common error is switching the order of subtraction or misplacing the negative sign for the $\hat{\jmath}$ component.
• Parameter Consistency: When finding the intersection of lines, always verify the calculated parameter values (e.g., $\delta$ and $\lambda$) by substituting them into all three coordinate equations. If they don't satisfy all three, the lines might be skew and not intersect.
• Foot of Perpendicular: Remember that the vector from the given point to a general point on the line must be perpendicular to the direction vector of the line, not necessarily the vector from the origin to the line's starting point.

4. Summary

The problem involved a sequence of 3D vector geometry concepts. First, we determined the direction of line $l$ using the cross product of the direction vectors of $l_1$ and $l_2$. Then, we found the point of intersection $P$ of lines $l$ and $l_1$ by equating their parametric forms. Finally, we calculated the foot of the perpendicular $Q$ from $P$ onto line $l_2$ by using the dot product property of perpendicular vectors. The coordinates of $Q$ were then used to find the required expression. The final computed value is 5.

The final answer is $\boxed{5}$.