Let → a = ^ i + ^ j + ^ k , → b and → c = ^ j - ^ k be three...

Given,

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

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

Here, \overset{\to}{a} \times \overset{\to}{c} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 1 & 1 \\ 0 & 1 & - 1\end{cases}\right| = - 2\overset{^}{i} + \overset{^}{j} + \overset{^}{k}

Projection of $\overset{\to}{b}$ on $\overset{\to}{a} \times \overset{\to}{c} = ℓ$

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

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

$∴ ℓ = \dfrac{2}{\sqrt{6}} ⇒ {ℓ}^{2} = \dfrac{4}{6}$

So, $3{ℓ}^{2} = 2$