Let $P$ be the foot of the perpendicular from the point $Q(10, -3, -1)$ on the line $\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z + 2}{-2}$. Then the area of the right-angled triangle $PQR$, where $R$ is the point $(3, -2, 1)$, is

Step 1: Find the coordinates of point P

First, we represent any point $P$ on the given line in terms of a parameter $\lambda$. The direction vector of the line is $\vec{d} = (7, -1, -2)$. Since $P$ is the foot of the perpendicular from $Q$ to the line, the vector $\vec{QP}$ must be perpendicular to the direction vector of the line. We set their dot product to zero to find the value of $\lambda$, and then substitute $\lambda$ back into the coordinates of $P$.

Step 2: Calculate the lengths of PQ and PR

Next, we calculate the lengths of the sides $PQ$ and $PR$ using the distance formula between two points. We express the coordinates of $Q$ and $R$ with a common denominator to simplify the calculations.

Step 3: Calculate the area of triangle PQR

The problem states that $PQR$ is a right-angled triangle. Since $P$ is the foot of the perpendicular from $Q$ to the line, the angle $\angle QPR$ is not necessarily $90^\circ$. We check if the right angle is at $Q$ by calculating the dot product of $\vec{QP}$ and $\vec{QR}$. Since $\vec{QP} \cdot \vec{QR} = 0$, the right angle is at $Q$. Therefore, the area of the triangle is $\frac{1}{2} \times PQ \times QR$. We calculate the length of $QR$ and then the area.

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