Answer
$\dfrac{1}{3}$
Work Step by Step
The cross product is defined as:
$u \times v=\begin{vmatrix}i&j&k\\m_1&m_2&m_3\\n_1&n_2&n_3\end{vmatrix}=\lt m_2n_3-m_3n_2, m_3n_1-m_1n_3, m_1n_2-m_2b_1 \gt$
Let us consider $\overrightarrow {a}=-j+k$, $\overrightarrow {b}=i+k$ and $\overrightarrow {c}=i-j$
Now,
$[\overrightarrow {a} \overrightarrow {b} \overrightarrow {c}]=\begin{vmatrix}0&-1&1\\1&0&1\\1&-1&0 \end{vmatrix}=-2$
Then, we have the volume $V$ can be found as:
$V=\dfrac{1}{6}|[\overrightarrow {a} \overrightarrow {b} \overrightarrow {c}]|=\dfrac{|-2|}{6}$
or, $V=\dfrac{2}{6}=\dfrac{1}{3}$
