Step-by-step Derivations

1. Identify Given Vectors We are given the diagonal vectors of a parallelogram PQRS: $\overrightarrow{\mathrm{PR}} = 3 \hat{i}+\hat{j}-2 \hat{k}$ $\overrightarrow{\mathrm{SQ}} = \hat{i}-3 \hat{j}-4 \hat{k}$ We are also given another vector: $\overrightarrow{\mathrm{PT}} = \hat{i}+2 \hat{j}+3 \hat{k}$

2. Relate Diagonals to Adjacent Sides Let the adjacent sides of the parallelogram originating from vertex P be $\overrightarrow{\mathrm{PQ}}$ and $\overrightarrow{\mathrm{PS}}$. Using the parallelogram law of vector addition, the sum and difference of the adjacent sides give the diagonal vectors. The diagonal starting from P is the sum of the adjacent sides: $\overrightarrow{\mathrm{PR}} = \overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{PS}} \quad ... (1)$ The other diagonal can be expressed as the difference of the sides: In triangle PQS, $\overrightarrow{\mathrm{PS}} + \overrightarrow{\mathrm{SQ}} = \overrightarrow{\mathrm{PQ}}$, which gives: $\overrightarrow{\mathrm{SQ}} = \overrightarrow{\mathrm{PQ}} - \overrightarrow{\mathrm{PS}} \quad ... (2)$

3. Solve for the Side Vectors We now have a system of two linear vector equations. We can solve for $\overrightarrow{\mathrm{PQ}}$ and $\overrightarrow{\mathrm{PS}}$.

   Adding equation (1) and (2): $(\overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{PS}}) + (\overrightarrow{\mathrm{PQ}} - \overrightarrow{\mathrm{PS}}) = \overrightarrow{\mathrm{PR}} + \overrightarrow{\mathrm{SQ}}$ $2 \overrightarrow{\mathrm{PQ}} = (3 \hat{i}+\hat{j}-2 \hat{k}) + (\hat{i}-3 \hat{j}-4 \hat{k})$ $2 \overrightarrow{\mathrm{PQ}} = (3+1)\hat{i} + (1-3)\hat{j} + (-2-4)\hat{k}$ $2 \overrightarrow{\mathrm{PQ}} = 4\hat{i} - 2\hat{j} - 6\hat{k}$ $\overrightarrow{\mathrm{PQ}} = 2\hat{i} - \hat{j} - 3\hat{k}$

   Subtracting equation (2) from (1): $(\overrightarrow{\mathrm{PQ}} + \overrightarrow{\mathrm{PS}}) - (\overrightarrow{\mathrm{PQ}} - \overrightarrow{\mathrm{PS}}) = \overrightarrow{\mathrm{PR}} - \overrightarrow{\mathrm{SQ}}$ $2 \overrightarrow{\mathrm{PS}} = (3 \hat{i}+\hat{j}-2 \hat{k}) - (\hat{i}-3 \hat{j}-4 \hat{k})$ $2 \overrightarrow{\mathrm{PS}} = (3-1)\hat{i} + (1-(-3))\hat{j} + (-2-(-4))\hat{k}$ $2 \overrightarrow{\mathrm{PS}} = 2\hat{i} + 4\hat{j} + 2\hat{k}$ $\overrightarrow{\mathrm{PS}} = \hat{i} + 2\hat{j} + \hat{k}$

4. Calculate the Volume of the Parallelepiped The volume $V$ of the parallelepiped determined by the coterminous vectors $\overrightarrow{\mathrm{PT}}$, $\overrightarrow{\mathrm{PQ}}$, and $\overrightarrow{\mathrm{PS}}$ is given by the magnitude of their scalar triple product: $V = |[\overrightarrow{\mathrm{PT}}, \overrightarrow{\mathrm{PQ}}, \overrightarrow{\mathrm{PS}}]| = |\overrightarrow{\mathrm{PT}} \cdot (\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PS}})|$ This can be calculated using the determinant of the components of the vectors: $V = \left| \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & -3 \\ 1 & 2 & 1 \end{vmatrix} \right|$ Evaluating the determinant: $V = |1((-1)(1) - (-3)(2)) - 2((2)(1) - (-3)(1)) + 3((2)(2) - (-1)(1))|$ $V = |1(-1 + 6) - 2(2 + 3) + 3(4 + 1)|$ $V = |1(5) - 2(5) + 3(5)|$ $V = |5 - 10 + 15|$ $V = |10| = 10$

5. Conclusion The volume of the parallelepiped is 10 cubic units. This corresponds to option C.

**Alternative Method:**The volume of the parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ where $\vec{b}$ and $\vec{c}$ are sides of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by $V = \frac{1}{2} |[\vec{a}, \vec{d_1}, \vec{d_2}]|$.Here, $\vec{a} = \overrightarrow{PT}$, $\vec{d_1} = \overrightarrow{PR}$, and $\vec{d_2} = \overrightarrow{SQ}$.$$ V = \frac{1}{2} |[\overrightarrow{PT}, \overrightarrow{PR}, \overrightarrow{SQ}]| = \frac{1}{2} \left| \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & -2 \\ 1 & -3 & -4 \end{vmatrix} \right| $$ $$ V = \frac{1}{2} |1(1(-4) - (-2)(-3)) - 2(3(-4) - (-2)(1)) + 3(3(-3) - 1(1))| $$ $$ V = \frac{1}{2} |1(-4 - 6) - 2(-12 + 2) + 3(-9 - 1)| $$ $$ V = \frac{1}{2} |1(-10) - 2(-10) + 3(-10)| $$ $$ V = \frac{1}{2} |-10 + 20 - 30| = \frac{1}{2} |-20| = \frac{1}{2}(20) = 10 $$ Both methods give the same result.