Let → a = 2 ^ i + 3 ^ j + 4 ^ k , → b = ^ i - 2 ^ j - 2 ^ k and → c...

Given that,

\overset{\to}{a} = 2\overset{^}{i} + 3\overset{^}{j} + 4\overset{^}{k}, \overset{\to}{b} = \overset{^}{i} - 2\overset{^}{j} - 2\overset{^}{k} and \overset{\to}{c} = - \overset{^}{i} + 4\overset{^}{j} + 3\overset{^}{k} .

Let us find $\overset{\to}{b} \times \overset{\to}{c}$

⇒ \overset{\to}{b} \times \overset{\to}{c} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 2 & - 2 \\ - 1 & 4 & 3\end{cases}\right|

= \left(- 6 + 8\right)\overset{^}{i} - \left(3 - 2\right)\overset{^}{j} + \left(4 - 2\right)\overset{^}{k}

⇒ \overset{\to}{b} \times \overset{\to}{c} = 2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}

Since $\overset{\to}{d}$ is perpendicular to both $\overset{\to}{b}\text{and}\overset{\to}{c}$

∴ \overset{\to}{d} = \lambda ( 2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k} )

Also $\overset{\to}{a} \cdot \overset{\to}{d} = 18$

⇒ \lambda\left(2\overset{^}{i} + 3\overset{^}{j} + 4\overset{^}{k}\right) . ( 2\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k} ) = 18

$⇒ 9\lambda = 18$

$⇒ \lambda = 2$

Now let us apply LaGrange's identity which is as follows,

$∴ | \overset{\to}{a} \times \overset{\to}{d}{|}^{2} = | \overset{\to}{a}{|}^{2} \cdot | \overset{\to}{d}{|}^{2} - ( \overset{\to}{a} \cdot \overset{\to}{d}{)}^{2}$

= {\left|2\overset{^}{i} + 3\overset{^}{j} + 4\overset{^}{k}\right|}^{2}{\left|4\overset{^}{i} - 2\overset{^}{j} + 4\overset{^}{k}\right|}^{2} - {\left(18\right)}^{2}

$= 720$