Here, $O A, O B, O C$ are the co-terminal edges of a rectangular parallelopiped of volume $V$.


Also, we know that the volume of rectangular parallelopiped $=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
ie, $\quad V=[\overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}}]$ $\ldots$ (i)
Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}$
then from figure
$\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{BP}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{CP}}=\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{a}}$
$\because$ (By vector addition)
Now, we find
$[\overrightarrow{\mathrm{AP}} \overrightarrow{\mathrm{BP}} \overrightarrow{\mathrm{CP}}]=[(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})(\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}})]$
$=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}})]$
$=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}]$
$[\because \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}=0]$
$\left\{\begin{array}{l}\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \\ \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{b}}\end{array}\right.$
$=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+0+\overrightarrow{\mathbf{b}}]$
$=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}+\overrightarrow{\mathrm{b}}]$
$=(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{c}}+(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{b}}$
$=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}}$
$=1+0-0-0+0+1$
$=2 \cdot 1$
$\because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}}=1$
$=2\{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})\}$
$=2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$=2[\overrightarrow{\mathrm{OA}} \overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OC}}]$
$=2 \mathrm{~V} \quad$ [from Eq. (i)]
Hence, $[\overrightarrow{\mathrm{AP}} \overrightarrow{\mathrm{BP}} \overrightarrow{\mathrm{CP}}]=2 V$