Let → a = ^ i + ^ j + ^ k and → b = ^ j - ^ k . If → c is a vector...

We know that, for vectors $\overset{\to}{a}, \overset{\to}{b}$ and $\overset{\to}{c}$
\overset{\to}{a} \cdot ( \overset{\to}{b} \times \overset{\to}{c} ) = \left[\begin{cases}\overset{\to}{a} & \overset{\to}{b} & \overset{\to}{c}\end{cases}\right] = \left[\begin{cases}\overset{\to}{c} & \overset{\to}{a} & \overset{\to}{b}\end{cases}\right]

$⇒ \overset{\to}{a} \cdot ( \overset{\to}{b} \times \overset{\to}{c} ) = ( \overset{\to}{c} \times \overset{\to}{a} ) \cdot \overset{\to}{b}$

$= - \overset{\to}{b} \cdot \overset{\to}{b}$

$∵ \overset{\to}{a} \times \overset{\to}{c} = \overset{\to}{b} ⇒ \overset{\to}{c} \times \overset{\to}{a} = - \overset{\to}{b}$

$= - {\left|\overset{\to}{b}\right|}^{2}$

$= - {\left(\sqrt{{1}^{2} + {\left(- 1\right)}^{2}}\right)}^{2}$

$= - 2$