1. Given vectors

$\vec a=(3,1,-1),\qquad \vec c=(2,-3,3)$

and

$\vec a=\vec b\times \vec c, \qquad |\vec b|^2=50.$

We need to find

$\left|72-\left|\vec b+\vec c\right|^2\right|.$

2. Use the property of cross product

Since

$\vec a=\vec b\times \vec c,$

we know that $\vec a$ is perpendicular to both $\vec b$ and $\vec c$.

So,

$\vec a\cdot \vec b=0, \qquad \vec a\cdot \vec c=0.$

Let us verify $\vec a\cdot \vec c$:

$\vec a\cdot \vec c=(3)(2)+(1)(-3)+(-1)(3)=6-3-3=0.$

So the condition is consistent.

3. Use the identity for magnitude of cross product

$|\vec a|^2=|\vec b\times \vec c|^2=|\vec b|^2|\vec c|^2-(\vec b\cdot \vec c)^2.$

First compute $|\vec a|^2$ and $|\vec c|^2$:

$|\vec a|^2=3^2+1^2+(-1)^2=9+1+1=11,$

$|\vec c|^2=2^2+(-3)^2+3^2=4+9+9=22.$

Also given:

$|\vec b|^2=50.$

Hence,

$11=50\cdot 22-(\vec b\cdot \vec c)^2.$

$11=1100-(\vec b\cdot \vec c)^2$

$(\vec b\cdot \vec c)^2=1089=33^2.$

Therefore,

$\vec b\cdot \vec c=\pm 33.$

4. Find $|\vec b+\vec c|^2$

Using

$|\vec b+\vec c|^2=|\vec b|^2+|\vec c|^2+2\vec b\cdot \vec c,$

we get

$|\vec b+\vec c|^2=50+22+2(\pm 33)=72\pm 66.$

So the two possible values are

$|\vec b+\vec c|^2=138 \quad \text{or} \quad 6.$

5. Evaluate the required expression

$\left|72-|\vec b+\vec c|^2\right|$

If $|\vec b+\vec c|^2=138$,

$|72-138|=66.$

If $|\vec b+\vec c|^2=6$,

$|72-6|=66.$

In both cases, the value is the same.

6. Final answer

$\boxed{66}$