Let → a = 3 ^ i + 2 ^ j + ^ k , → b = 2 ^ i - ^ j + 3 ^ k and → c...

Given,

\overset{\to}{a} = 3\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}, \overset{\to}{b} = 2\overset{^}{i} - \overset{^}{j} + 3\overset{^}{k} and\overset{\to}{c}be a vector such that \left(\overset{\to}{a} + \overset{\to}{b}\right) \times \overset{\to}{c} = 2\left(\overset{\to}{a} \times \overset{\to}{b}\right) + 24\overset{^}{j} - 6\overset{^}{k} 

Now, finding \overset{\to}{a} \times \overset{\to}{b} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 3 & 2 & 1 \\ 2 & - 1 & 3\end{cases}\right| = 7\overset{^}{i} - 7\overset{^}{j} - 7\overset{^}{k}

Now, solving

\left(\overset{\to}{a} + \overset{\to}{b}\right) \times \overset{\to}{c} = 2\left(\overset{\to}{a} \times \overset{\to}{b}\right) + 24\overset{^}{j} - 6\overset{^}{k}$</p><p>$⇒ \left(5\overset{^}{i} + \overset{^}{j} + 4\overset{^}{k}\right) \times \overset{\to}{c} = 2\left(7\overset{^}{i} - 7\overset{^}{j} - 7\overset{^}{k}\right) + 24\overset{^}{j} - 6\overset{^}{k}$</p><p>$⇒ \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 5 & 1 & 4 \\ x & y & z\end{cases}\right| = 14\overset{^}{i} + 10\overset{^}{j} - 20\overset{^}{k}$</p><p>$⇒ \overset{^}{i}\left(z - 4y\right) - \overset{^}{j}\left(5z - 4x\right) + \overset{^}{k}\left(5y - x\right) = 14\overset{^}{i} + 10\overset{^}{j} - 20\overset{^}{k}$</p><p>Now, on comparing both side, </p><p>$z - 4y = 14, 4x - 5z = 10, 5y - x = - 20 . . . . \left(i\right)</p><p>Now, solving </p><p>\left(\overset{\to}{a} - \overset{\to}{b} + \overset{^}{i}\right) \cdot \overset{\to}{c} = - 3

⇒ \left(2\overset{^}{i} + 3\overset{^}{j} - 2\overset{^}{k}\right) \cdot \overset{\to}{c} = - 3

$⇒ 2x + 3y - 2z = - 3 . . . . . . . \left(ii\right)$

Now, solving the equation \left(i\right) \text{ and } \left(ii\right) we get,

$∴ x = 5, y = - 3, z = 2$

$⇒ {\left|\overset{\to}{c}\right|}$^{2} = 25 + 9 + 4 = 38$</p>$