If → a = ^ i + 2 ^ j + ^ k , → b = 3 ( ^ i - ^ j + ^ k ) and → c be...

Given: \overset{\to}{a} = \overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}, \overset{\to}{b} = 3\left(\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right), $\overset{\to}{a} \cdot \overset{\to}{c} = 3$ and $\overset{\to}{a} \times \overset{\to}{c} = \overset{\to}{b}$.

Now, $\overset{\to}{a} \times \overset{\to}{c} = \overset{\to}{b}$

$⇒ \left(\overset{\to}{a} \times \overset{\to}{c}\right) \cdot \overset{\to}{b} = {\left|\overset{\to}{b}\right|}^{2}$

$⇒ \left(\overset{\to}{a} \times \overset{\to}{c}\right) \cdot \overset{\to}{b} = {\left(3\sqrt{{1}^{2} + {1}^{2} + {1}^{2}}\right)}^{2}$

$⇒ \left(\overset{\to}{a} \times \overset{\to}{c}\right) \cdot \overset{\to}{b} = 27$

$⇒ \overset{\to}{a} \cdot \left(\overset{\to}{c} \times \overset{\to}{b}\right) = 27 . . . \left(i\right)$

\left\{\text{as}\overset{\to}{a} \cdot \left(\overset{\to}{c} \times \overset{\to}{b}\right) = \overset{\to}{b} \cdot \left(\overset{\to}{a} \times \overset{\to}{c}\right)\right\}$</p><p>Also,$\overset{\to}{a} \cdot \overset{\to}{b} = 3 - 6 + 3 = 0$</p><p>$⇒ \overset{\to}{a} \cdot \left(\left(\overset{\to}{c} \times \overset{\to}{b}\right) - \overset{\to}{b} - \overset{\to}{c}\right) = \overset{\to}{a} \cdot \left(\overset{\to}{c} \times \overset{\to}{b}\right) - \overset{\to}{a} \cdot \overset{\to}{b} - \overset{\to}{a} \cdot \overset{\to}{c}$</p><p>$⇒ \overset{\to}{a} \cdot \left(\left(\overset{\to}{c} \times \overset{\to}{b}\right) - \overset{\to}{b} - \overset{\to}{c}\right) = 27 - 0 - 3$</p><p>$⇒ \overset{\to}{a} \cdot \left(\left(\overset{\to}{c} \times \overset{\to}{b}\right) - \overset{\to}{b} - \overset{\to}{c}\right) = 24$