1. Key Concepts and Formulas

• Equation of a Line in 3D: A line passing through a point $(x_1, y_1, z_1)$ and parallel to a direction vector $\vec{d} = \langle a, b, c \rangle$ can be expressed in symmetric form as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. This can be parameterized by setting each ratio equal to a variable, say $\lambda$, yielding $x = x_1 + a\lambda$, $y = y_1 + b\lambda$, $z = z_1 + c\lambda$.
• Interpreting "Distance Along a Line": When asked for the distance of a point $P$ from a line $L_1$ along another line $L_2$, it signifies finding a point $Q$ on $L_1$ such that the line segment $PQ$ is parallel to $L_2$. The desired distance is the length of this segment $PQ$. This is distinct from the perpendicular distance.
• Distance Formula in 3D: The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

2. Step-by-Step Solution

Let the given point be $P(7,10,11)$. Let the target line be $L_1: \frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$. Let the direction-specifying line be $L_2: \frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$.

Step 1: Determine the Direction Vector for the Search Line

• What we are doing: We need to find the direction in which we "search" for the point on $L_1$. The problem specifies this as "along the line $L_2$".
• Why we are doing this: This means the line segment connecting point $P$ to the target line $L_1$ must be parallel to $L_2$. Therefore, the direction vector of this segment (and the line containing it) will be the direction vector of $L_2$.
• From the symmetric equation of $L_2$, $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$, the direction vector is $\vec{d} = \langle 2, 3, 6 \rangle$.

Step 2: Formulate the Equation of the Line Passing Through $P$ in the Desired Direction

• What we are doing: We construct a new line, let's call it $L_P$, which passes through the given point $P(7,10,11)$ and is parallel to the direction $\vec{d} = \langle 2, 3, 6 \rangle$ identified in Step 1.
• Why we are doing this: The point $Q$ we are looking for lies on $L_1$ and also on $L_P$. Finding the intersection of $L_P$ and $L_1$ will give us point $Q$.
• Using point $P(7,10,11)$ and the direction vector $\vec{d} = \langle 2, 3, 6 \rangle$, the symmetric equation of line $L_P$ is: $\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}$
• To represent any point on this line $L_P$, we introduce a parameter $\lambda$: $\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda$
• Thus, any point $Q$ on $L_P$ can be written in parametric form as: $Q(x,y,z) = (2\lambda+7, 3\lambda+10, 6\lambda+11)$

Step 3: Find the Point of Intersection $Q$ with the Target Line $L_1$

• What we are doing: We find the specific point $Q$ that lies on both $L_P$ (the line we just constructed) and $L_1$ (the given target line).
• Why we are doing this: This intersection point $Q$ is the unique point on $L_1$ such that the segment $PQ$ is parallel to $L_2$, as required by the problem statement.
• The point $Q(2\lambda+7, 3\lambda+10, 6\lambda+11)$ must satisfy the equation of line $L_1: \frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$.
• Important consideration for $L_1$: When a denominator in the symmetric form of a line equation is zero (e.g., $\frac{y-4}{0}$), it implies that the corresponding numerator must also be zero for the line to be defined, and that coordinate is constant for all points on the line. So, for $L_1$, $\frac{y-4}{0}$ means $y-4=0$, which simplifies to $y=4$.
• Substitute the $y$-coordinate of $Q$ into this condition for $L_1$: $3\lambda+10 = 4$ $3\lambda = 4 - 10$ $3\lambda = -6$ $\lambda = -2$
• Now, we substitute $\lambda=-2$ back into the parametric coordinates of $Q$ to find its specific coordinates: $Q(x,y,z) = (2(-2)+7, 3(-2)+10, 6(-2)+11)$ $Q(x,y,z) = (-4+7, -6+10, -12+11)$ $Q(x,y,z) = (3, 4, -1)$
• We can quickly verify these coordinates satisfy the other parts of $L_1$: $\frac{3-4}{1} = -1$ and $\frac{-1-2}{3} = \frac{-3}{3} = -1$. All components are consistent.

Step 4: Calculate the Distance Between $P$ and $Q$

• What we are doing: We have found the initial point $P(7,10,11)$ and the final point $Q(3,4,-1)$ of the line segment whose length represents the required distance.
• Why we are doing this: This is the direct application of the 3D distance formula to find the length of the segment $PQ$.
• Using the distance formula for $P(7,10,11)$ and $Q(3,4,-1)$: $PQ = \sqrt{(3-7)^2 + (4-10)^2 + (-1-11)^2}$ $PQ = \sqrt{(-4)^2 + (-6)^2 + (-12)^2}$ $PQ = \sqrt{16 + 36 + 144}$ $PQ = \sqrt{196}$ $PQ = 14$

3. Common Mistakes & Tips

• Misinterpreting "Along the Line": A frequent error is to confuse "distance along a line" with the perpendicular distance from a point to a line. Always remember that "along the line" implies parallelism to the specified direction.
• Handling Zero Denominators: Be meticulous when dealing with a zero in the denominator of a line's symmetric equation (e.g., $\frac{y-4}{0}$). This does not mean the line is undefined; it means the corresponding coordinate is constant (e.g., $y=4$).
• Algebraic Errors: Carefully perform arithmetic operations, especially squaring negative numbers and summing terms under the square root, as small calculation mistakes can lead to an incorrect final answer.

4. Summary

This problem requires understanding how to find the distance of a point from a line along a specific direction. The strategy involves constructing a new line that passes through the given point and is parallel to the specified direction line. The intersection of this new line with the target line gives the endpoint of the segment whose length is the required distance. Finally, the 3D distance formula is applied to calculate this length. This systematic approach ensures all conditions of the problem are met.

The final answer is $\boxed{\text{14}}$, which corresponds to option (A).