This solution will guide you through finding the intersection point of two lines and then calculating the distance from that point to a third line using vector methods.

1. Key Concepts and Formulas

   • Parametric Form of a Line in 3D: A line passing through a point $(x_0, y_0, z_0)$ with a direction vector $\vec{d} = (a, b, c)$ can be represented in parametric form as: $x = x_0 + at$ $y = y_0 + bt$ $z = z_0 + ct$ where $t$ is a parameter. This form is essential for finding points on the line and for determining the intersection of lines.

   • Intersection of Two Lines: Two lines intersect if there exists a common point that lies on both lines. To find this point, we express both lines in their parametric forms (using different parameters) and then equate their corresponding $x, y, z$ coordinates. Solving the resulting system of linear equations for the parameters will yield the values that correspond to the intersection point. A crucial step is to check if these parameter values are consistent across all three coordinate equations.

   • Distance of a Point from a Line: The shortest distance $l$ from a point $P$ to a line passing through a point $A$ with a direction vector $\vec{d}$ can be calculated using vector projection. Let $\vec{AP}$ be the vector from point $A$ on the line to point $P$. The square of the distance $l$ is given by: $l^2 = |\vec{AP}|^2 - \left(\frac{\vec{AP} \cdot \vec{d}}{|\vec{d}|}\right)^2$ This formula stems from the Pythagorean theorem, where the distance $l$ is one leg of a right triangle, $|\vec{AP}|$ is the hypotenuse, and the projection of $\vec{AP}$ onto $\vec{d}$ is the other leg. Alternatively, the distance can be found using the cross product: $l = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$. Both methods yield the same result.

2. Step-by-Step Solution

   Step 1: Represent the given lines in parametric form. We are given two lines, $L_1$ and $L_2$, in symmetric form. To find their intersection point, we convert them into parametric form. This allows us to represent any point on each line using a single variable.

   • For Line $L_1$: $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16} = k$ From this, we can express the coordinates of any point on $L_1$ in terms of parameter $k$: $x = 2k+2$ $y = -2k$ $z = 16k+7$ So, any point on $L_1$ is $(2k+2, -2k, 16k+7)$.

   • For Line $L_2$: $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1} = \mu$ Similarly, for $L_2$ using parameter $\mu$: $x = 4\mu-3$ $y = 3\mu-2$ $z = \mu-2$ Any point on $L_2$ is $(4\mu-3, 3\mu-2, \mu-2)$.

   Step 2: Find the intersection point P by equating coordinates. If the lines intersect, there must be a common point $(x, y, z)$ that satisfies both sets of parametric equations for specific values of $k$ and $\mu$. We equate the corresponding coordinates:

   1. Equating $x$-coordinates: $2k+2 = 4\mu-3 \implies 2k - 4\mu = -5 \quad \text{(Equation 1)}$

   2. Equating $y$-coordinates: $-2k = 3\mu-2 \implies -2k - 3\mu = -2 \quad \text{(Equation 2)}$

   3. Equating $z$-coordinates: $16k+7 = \mu-2 \implies 16k - \mu = -9 \quad \text{(Equation 3)}$

   Now, we solve the system of linear equations for $k$ and $\mu$. We can use Equations 1 and 2: Adding (Equation 1) and (Equation 2) eliminates $k$: $(2k - 4\mu) + (-2k - 3\mu) = -5 + (-2)$ $-7\mu = -7$ $\mu = 1$ Substitute $\mu=1$ into Equation 2 to find $k$: $-2k - 3(1) = -2$ $-2k - 3 = -2$ $-2k = 1$ $k = -\frac{1}{2}$

   Step 3: Verify the parameters and determine the intersection point P. It is crucial to verify if the values of $k$ and $\mu$ obtained from two equations satisfy the third equation. If they do, the lines intersect at a unique point; otherwise, the lines are skew. Substitute $k = -1/2$ and $\mu = 1$ into Equation 3: $16\left(-\frac{1}{2}\right) - 1 = -8 - 1 = -9$ Since this matches the right-hand side of Equation 3, the values are consistent, and the lines indeed intersect.

   Now, substitute $k = -1/2$ into the parametric equations for $L_1$ to find the coordinates of point $P$: $P_x = 2\left(-\frac{1}{2}\right)+2 = -1+2 = 1$ $P_y = -2\left(-\frac{1}{2}\right) = 1$ $P_z = 16\left(-\frac{1}{2}\right)+7 = -8+7 = -1$ Thus, the intersection point $P$ is $(1, 1, -1)$. (Alternatively, substituting $\mu=1$ into $L_2$'s equations yields the same point $P=(1,1,-1)$).

   Step 4: Identify the components of the third line ($L_3$). We need to find the distance of $P(1, 1, -1)$ from the line $L_3$: $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ From the symmetric form of $L_3$, we can identify:

   • A point on the line $A$: Comparing with $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$, we get $A = (-1, 1, 1)$.
   • The direction vector of the line $\vec{d}$: The denominators give the direction ratios, so $\vec{d} = (2, 3, 1)$.

   Step 5: Calculate the vector $\vec{AP}$ and its magnitude, and the dot product with $\vec{d}$. First, form the vector $\vec{AP}$ from point $A$ on $L_3$ to point $P$: $\vec{AP} = P - A = (1 - (-1), 1 - 1, -1 - 1) = (2, 0, -2)$ Next, calculate the square of the magnitude of $\vec{AP}$ and $\vec{d}$: $|\vec{AP}|^2 = 2^2 + 0^2 + (-2)^2 = 4 + 0 + 4 = 8$ $|\vec{d}|^2 = 2^2 + 3^2 + 1^2 = 4 + 9 + 1 = 14$ Finally, calculate the dot product of $\vec{AP}$ and $\vec{d}$: $\vec{AP} \cdot \vec{d} = (2)(2) + (0)(3) + (-2)(1) = 4 + 0 - 2 = 2$

   Step 6: Apply the distance formula to find $l^2$. Using the distance formula $l^2 = |\vec{AP}|^2 - \frac{(\vec{AP} \cdot \vec{d})^2}{|\vec{d}|^2}$: $l^2 = 8 - \frac{(2)^2}{14}$ $l^2 = 8 - \frac{4}{14}$ $l^2 = 8 - \frac{2}{7}$ To combine these terms, find a common denominator: $l^2 = \frac{8 \times 7}{7} - \frac{2}{7} = \frac{56 - 2}{7} = \frac{54}{7}$

   Step 7: Calculate the final required value $14l^2$. The problem asks for the value of $14l^2$. $14l^2 = 14 \times \frac{54}{7}$ $14l^2 = 2 \times 54$ $14l^2 = 108$

3. Common Mistakes & Tips

   • Using the same parameter: When finding the intersection of two lines, always use distinct parameters (e.g., $k$ and $\mu$) for each line. Using the same parameter implies the lines are identical or parallel, which is generally not the case for intersecting or skew lines.
   • Consistency Check: After finding the values of the parameters from two equations, always substitute them into the third equation to verify consistency. If the third equation is not satisfied, the lines are skew and do not intersect.
   • Arithmetic Errors: 3D geometry problems often involve several calculations with integers and fractions. Double-check each step, especially when solving systems of equations, performing vector operations (dot products, cross products), and simplifying fractions.
   • Correctly Identifying Components: Ensure you correctly identify the point on the line $(x_1, y_1, z_1)$ and the direction vector components $(a, b, c)$ from the symmetric form. Remember that $x+a$ implies $x_1 = -a$.

4. Summary

   To solve this problem, we first converted the equations of the two intersecting lines into their parametric forms. By equating the corresponding coordinates, we formed a system of linear equations, which we solved to find the parameters $k$ and $\mu$. A consistency check confirmed the intersection, allowing us to determine the coordinates of point $P$. Next, we identified a point and the direction vector of the third line. Using the vector projection formula, we calculated the square of the distance ($l^2$) from point $P$ to this third line. Finally, we computed the required value of $14l^2$.

The final answer is $\boxed{108}$.