Given that,

$\overrightarrow{a}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}, \overrightarrow{b}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}-2\hat{\mathrm{k}}$ and $\overrightarrow{c}=-\hat{\mathrm{i}}+4\hat{\mathrm{j}}+3\hat{\mathrm{k}}.$

Let us find $\overrightarrow{b}\times \overrightarrow{c}$

$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& -2& -2\\ -1& 4& 3\end{array}\right|$

$=\left(-6+8\right)\hat{\mathrm{i}}-\left(3-2\right)\hat{\mathrm{j}}+\left(4-2\right)\hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

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

$\therefore \overrightarrow{d}=\lambda (2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})$

Also $\overrightarrow{\mathit{a}}\mathit{·}\overrightarrow{\mathit{d}}=18$

$\Rightarrow \lambda \left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right).(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}})=18$

$\Rightarrow 9\lambda =18$

$\Rightarrow \lambda =2$

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

$\therefore |\overrightarrow{a}\times \overrightarrow{d}{|}^2=|\overrightarrow{a}{|}^2·|\overrightarrow{d}{|}^2-(\overrightarrow{a}·\overrightarrow{d}{)}^2$

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

$=720$