Given,

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

And $\overrightarrow{b}$ is a vector such that $\overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{c}$ and $|\overrightarrow{b}{|}^2=50,$

Now solving, $\left|\overrightarrow{a}\right|=|\overrightarrow{b}\times \overrightarrow{c}|$

$\Rightarrow \sqrt{3^2+1^2+1^2}=\left|\overrightarrow{b}\right|·\sqrt{2^2+3^2+3^2}·\sin \theta$

$\Rightarrow \sqrt{11}=\sqrt{50}·\sqrt{22}·\sin \theta$

$\therefore \sin \theta =\frac{1}{10}$ or $\cos \theta =\frac{\sqrt{99}}{10}$

Now solving,$\left|72-{\left|\overrightarrow{b}+\overrightarrow{c}\right|}^2\right|$ we get,

$\left|72-{\left|\overrightarrow{b}+\overrightarrow{c}\right|}^2\right|$

$=\left|72-\left({\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos \theta \right)\right|$

$=\left|72-\left(50+22+2\times 5\sqrt{2}.\sqrt{22}\frac{\sqrt{99}}{10}\right)\right|$

$=\left|72-\left(72+\frac{2\times 5\times 2\times 11\times 3}{10}\right.\right|$

$=\left|66\right|=66$