Given the tethraedron $OABC$ where $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$ and $\vec{OC} = \vec{c}$ and the points $M$, $N$ and $P$, which are the midpoints of segments $\vec{AC}$, $\vec{AB}$ and $\vec{BC}$ respectively, find the ratio between $\vert \vec{OM} \cdot \vec{ON} \times \vec{OP} \vert $ and the volume of the tethraedron $V$.
My approach was:
$$ \vec{OM} = \vec{OA} + \frac{1}{2} \vec{AC} = \vec{a} + \frac{1}{2} (- \vec{a} + \vec{c}) = \frac{\vec{a}}{2} + \frac{\vec{c}}{2}\\ \vec{ON} = \cdots = \frac{\vec{a}}{2} + \frac{\vec{b}}{2}\\ \vec{OP} = \cdots = \frac{\vec{c}}{2} + \frac{\vec{b}}{2}\\ $$
$$ \begin{align*} \vert \vec{OM} \cdot \vec{ON} \times \vec{OP} \vert &= \left\vert \left( \frac{\vec{a}}{2} + \frac{\vec{c}}{2} \right) \cdot \left( \frac{\vec{a}}{2} + \frac{\vec{b}}{2} \right) \times \left( \frac{\vec{c}}{2} + \frac{\vec{b}}{2} \right) \right\vert\\ &= \frac{1}{8} \left\vert \left( \vec{a} + \vec{c} \right) \cdot \left( \vec{a} + \vec{b} \right) \times \left( \vec{c} + \vec{b} \right) \right\vert\\ &= \frac{1}{8} \left\vert (\vec{a} + \vec{c}) \cdot (\vec{a} \times \vec{c} + \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + 0) \right\vert \\ &= \frac{1}{8} \left\vert \vec{a} \cdot \vec{a} \times \vec{c} + \vec{a} \cdot \vec{a} \times \vec{b} + \vec{a} \cdot \vec{b} \times \vec{c} + \vec{c} \cdot \vec{a} \times \vec{c} + \vec{c} \cdot \vec{a} \times \vec{b} + \vec{c} \cdot \vec{b} \times \vec{c} \right\vert\\ &= \frac{1}{8} \left\vert 2 \vec{a} \cdot \vec{b} \times \vec{c} \right\vert\\ &= \frac{1}{4} \left\vert \vec{a} \cdot \vec{b} \times \vec{c} \right\vert \end{align*} $$
Since
$$V = \frac{1}{6} \left\vert \vec{a} \cdot \vec{b} \times \vec{c} \right\vert $$
then
$$ \frac{\vert \vec{OM} \cdot \vec{ON} \times \vec{OP} \vert }{V} = \frac{\frac{1}{4}}{\frac{1}{6}} = \frac{3}{2} $$
Did I make a mistake somewhere?
Textbook's answer: $\vert \vec{OM} \cdot \vec{ON} \times \vec{OP} \vert = \frac{3}{4} V$
Let $[X]$ denote the volume of a $3$-dimensional object $X$, and $\langle Y\rangle$ the area of a $2$-dimensional object $Y$. Then, $$[OMNP]=\frac{1}{6}\Biggl|\overrightarrow{OM}\cdot\Big(\overrightarrow{ON}\times\overrightarrow{OP}\Big)\Biggr|\,.$$ Note that the tetrahedra $OMNP$ and $OABC$ share a common altitude (from the point $O$). Since $MNP$ and $ABC$ are similar triangles with $\frac{MN}{BC}=\frac{NP}{CA}=\frac{PM}{AB}=\frac{1}{2}$, we get $$\frac{[OMNP]}{[OABC]}=\frac{\langle MNP\rangle }{\langle ABC\rangle}=\left(\frac{1}{2}\right)^2=\frac14\,,$$ That is, $$\frac{\Biggl|\overrightarrow{OM}\cdot\Big(\overrightarrow{ON}\times\overrightarrow{OP}\Big)\Biggr|}{V}=6\left(\frac{[OMNP]}{[OABC]}\right)=\frac{6}{4}=\frac{3}{2}\,.$$ So, you are correct, and the mistake is in the textbook.