Let $\overrightarrow{\text{PR}}=3\widehat{i}+\widehat{j}-2\widehat{k}\text{and}\overrightarrow{\text{SQ}}=\widehat{i}-3\widehat{j}-4\widehat{k}$ determine diagonals of a parallelogram PQRS and $\overrightarrow{\text{PT}}=\widehat{i}+2\widehat{j}+3\widehat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{\mathit{\text{PT}}}\text{,}\overrightarrow{\mathit{\text{PQ}}}\text{and}\overrightarrow{\mathit{\text{PS}}}$  is

• A

  5
• B

  20
• Correct answer
  C

  10
• D

  30

Solution

$\overrightarrow{\text{PR}}=\overrightarrow{\text{PQ}}+\overrightarrow{\text{PS}}$
$\overrightarrow{\text{SQ}}=\overrightarrow{\text{PQ}}-\overrightarrow{\text{PS}}$
$\overrightarrow{\text{PS}}=\frac{\overrightarrow{\text{PR}}-\overrightarrow{\text{SQ}}}{2}$

$\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PR}}+\overrightarrow{\text{SQ}}}{2}$
$\text{V}=\left|\left[\overrightarrow{\text{PQ}}\overrightarrow{\text{PS}}\overrightarrow{\text{PT}}\right]\right|$
$\text{V}=\frac{1}{4}\left|\left[\overrightarrow{\text{PR}}+\overrightarrow{\text{SQ}}\text{, }\overrightarrow{\text{PR}}-\overrightarrow{\text{SQ}}\text{, }\overrightarrow{\text{PT}}\right]\right|$
$\text{V}=\frac{1}{2}\left|\left[\overrightarrow{\text{PR}}\text{, }\overrightarrow{\text{SQ}}\text{, }\overrightarrow{\text{PT}}\right]\right|$
$V=\frac{1}{2}\left|\begin{array}{ccc}3& 1& -2\\ 1& -3& -4\\ 1& 2& 3\end{array}\right|$
$V=\frac{1}{2}\left(-3-7-10\right)=10$

Hence, the correct answer is option: 'C'.