1. Key Concepts and Formulas

This problem requires the application of several fundamental concepts from 3D Geometry to determine a distance between two specific points. The essential tools we will use are:

• Centroid of a Triangle: The centroid ($G$) of a triangle is the point of intersection of its medians. It divides each median in a $2:1$ ratio, measured from the vertex to the midpoint of the opposite side. If $R(x_R, y_R, z_R)$ is a vertex and $M(x_M, y_M, z_M)$ is the midpoint of the side opposite to $R$, the coordinates of the centroid $G$ are given by the section formula: $G = \left( \frac{2x_M + 1x_R}{2+1}, \frac{2y_M + 1y_R}{2+1}, \frac{2z_M + 1z_R}{2+1} \right)$
• Equation of a Line in Parametric Form: A line given in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ can be expressed in parametric form by setting each ratio equal to a scalar parameter, say $\lambda$. This yields general coordinates for any point on the line: $(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$. If a direction ratio is zero (e.g., $a=0$), then $x-x_1 = 0 \cdot \lambda \Rightarrow x=x_1$, meaning the coordinate is constant.
• Intersection of Two Lines: To find the point where two lines intersect, we represent a general point on each line using its parametric form (with different parameters). By equating the corresponding $x, y, z$ coordinates, we form a system of equations. Solving this system for the parameters will give the coordinates of the intersection point, provided the lines are not parallel and indeed intersect. A consistency check using the third coordinate is crucial.
• Distance Formula in 3D: The distance ($D$) between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in three dimensions is calculated as: $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

2. Step-by-Step Solution

Step 1: Determine the Coordinates of the Centroid ($G$) of $\triangle PQR$.

• Why this step? The problem asks for the distance of the centroid from another point. Therefore, the first logical step is to calculate the coordinates of the centroid.
• We are given the coordinates of vertex $R(-1,4,2)$ and the midpoint of the side $PQ$, which is $M(2,1,2)$.
• In $\triangle PQR$, the line segment $RM$ is a median. The centroid $G$ divides the median $RM$ in the ratio $2:1$, where the ratio is taken from the vertex $R$ to the midpoint $M$. This means $G$ divides $RM$ such that $RG:GM = 2:1$. However, the standard section formula for a centroid using a vertex and the midpoint of the opposite side considers the ratio $2:1$ from the midpoint to the vertex, or $1:2$ from the vertex to the midpoint, or more simply, it's the average of the coordinates if we consider all vertices, or a weighted average if we consider the vertex and the centroid. A more direct application of the section formula for a point dividing a line segment $AB$ in ratio $m:n$ is $\frac{n A + m B}{m+n}$. Here, $G$ divides $RM$ in the ratio $1:2$ from $R$ to $M$, or $2:1$ from $M$ to $R$. Using the coordinates of $R(x_R, y_R, z_R) = (-1,4,2)$ and $M(x_M, y_M, z_M) = (2,1,2)$, and considering $G$ divides $RM$ such that $G$ is $2/3$ of the way from $R$ to the midpoint of $PQ$, $M$. A simpler way to think about it is that the centroid $G$ is given by $G = \frac{R + P + Q}{3}$. Since $M = \frac{P+Q}{2}$, then $P+Q = 2M$. So, $G = \frac{R + 2M}{3}$. This is equivalent to applying the section formula for $G$ dividing $RM$ in the ratio $2:1$ from $M$ to $R$.
• Using the formula $G = \left( \frac{1x_R + 2x_M}{1+2}, \frac{1y_R + 2y_M}{1+2}, \frac{1z_R + 2z_M}{1+2} \right)$: $G_x = \frac{1(-1) + 2(2)}{3} = \frac{-1+4}{3} = \frac{3}{3} = 1$ $G_y = \frac{1(4) + 2(1)}{3} = \frac{4+2}{3} = \frac{6}{3} = 2$ $G_z = \frac{1(2) + 2(2)}{3} = \frac{2+4}{3} = \frac{6}{3} = 2$
• Thus, the coordinates of the centroid of $\triangle PQR$ are $G(1,2,2)$.

Step 2: Find the Point of Intersection ($A$) of the Two Given Lines.

• Why this step? The problem requires the distance from the centroid to the point of intersection of two lines. We must first find the coordinates of this intersection point.

• Let the first line be $L_1$: $\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$.

  • We express a general point on $L_1$ using a parameter, say $\lambda$: $\frac{x-2}{0} = \lambda \Rightarrow x-2 = 0 \cdot \lambda \Rightarrow x = 2$ (Since the direction ratio for $x$ is 0, the $x$-coordinate is constant for any point on this line). $\frac{y}{2} = \lambda \Rightarrow y = 2\lambda$ $\frac{z+3}{-1} = \lambda \Rightarrow z = -\lambda - 3$
  • So, any point on $L_1$ can be represented as $(2, 2\lambda, -3-\lambda)$.

• Let the second line be $L_2$: $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$.

  • We express a general point on $L_2$ using a different parameter, say $\mu$: $\frac{x-1}{1} = \mu \Rightarrow x = \mu + 1$ $\frac{y+3}{-3} = \mu \Rightarrow y = -3\mu - 3$ $\frac{z+1}{1} = \mu \Rightarrow z = \mu - 1$
  • So, any point on $L_2$ can be represented as $(\mu+1, -3\mu-3, \mu-1)$.

• For the lines to intersect, their coordinates must be equal at some specific values of $\lambda$ and $\mu$. Equating the corresponding coordinates:

  1. Equating $x$-coordinates: $2 = \mu + 1 \Rightarrow \mu = 1$.
  2. Equating $y$-coordinates: $2\lambda = -3\mu - 3$. Substitute the value of $\mu=1$: $2\lambda = -3(1) - 3 \Rightarrow 2\lambda = -6 \Rightarrow \lambda = -3$.
  3. Equating $z$-coordinates (Consistency Check): Now, we must verify if these values of $\lambda$ and $\mu$ are consistent for the $z$-coordinates. For $L_1$ (using $\lambda=-3$): $z = -\lambda - 3 = -(-3) - 3 = 3 - 3 = 0$. For $L_2$ (using $\mu=1$): $z = \mu - 1 = (1) - 1 = 0$.
  • Since the $z$-coordinates match, the lines intersect at a unique point.

• Substitute $\lambda=-3$ into the general point for $L_1$ (or $\mu=1$ into $L_2$) to find the intersection point $A$: Using $L_1$: $A(2, 2(-3), -3-(-3)) = A(2, -6, 0)$.

Step 3: Calculate the Distance Between the Centroid ($G$) and the Point of Intersection ($A$).

• Why this step? This is the final calculation required to answer the problem, as it asks for the distance between the two points we just found.
• We have the centroid $G(1,2,2)$ and the point of intersection $A(2,-6,0)$.
• Using the 3D distance formula: $AG = \sqrt{(x_A - x_G)^2 + (y_A - y_G)^2 + (z_A - z_G)^2}$ $AG = \sqrt{(2-1)^2 + (-6-2)^2 + (0-2)^2}$ $AG = \sqrt{(1)^2 + (-8)^2 + (-2)^2}$ $AG = \sqrt{1 + 64 + 4}$ $AG = \sqrt{69}$

3. Common Mistakes & Tips

• Centroid Formula: Be careful with the section formula for the centroid. If you use a vertex $R$ and the midpoint $M$ of the opposite side, the centroid $G$ divides $RM$ in the ratio $2:1$ from $R$ to $M$ (meaning $RG:GM = 2:1$). Or, more robustly, $G = \frac{R + P + Q}{3}$ and if $M$ is midpoint of $PQ$, $M = \frac{P+Q}{2} \Rightarrow P+Q = 2M$. So $G = \frac{R+2M}{3}$.
• Zero Direction Ratios: When a denominator in the symmetric form of a line is zero (e.g., $\frac{x-x_1}{0}$), it implies that the numerator must also be zero for the line to be well-defined. This means the corresponding coordinate is constant (e.g., $x=x_1$). Treat this carefully when converting to parametric form.
• Consistency Check: After finding the values of the parameters (e.g., $\lambda$ and $\mu$) from two coordinate equations (e.g., $x$ and $y$), always substitute them back into the third coordinate equation (e.g., $z$) for both lines to ensure they yield the same value. If they do not match, the lines are skew and do not intersect.

4. Summary

This problem required a systematic approach involving fundamental concepts of 3D geometry. We first calculated the coordinates of the centroid of $\triangle PQR$ using the given vertex $R$ and the midpoint $M$ of the side $PQ$. Next, we found the point of intersection of the two given lines by converting their equations into parametric form, equating coordinates, and solving for the parameters. Finally, we applied the 3D distance formula to determine the distance between the calculated centroid and the intersection point. The result obtained was $\sqrt{69}$.

5. Final Answer

The final answer is $\boxed{\sqrt{69}}$, which corresponds to option (C).