1. Since $\vec u$ is coplanar with $\vec a$ and $\vec b$, write $\vec u = x\vec a + y\vec b$ for some scalars $x,y$.

Given $\vec a=(2,3,-1),\qquad \vec b=(0,1,1).$ So, $\vec u=x(2,3,-1)+y(0,1,1)=(2x,3x+y,-x+y).$

2. Use the condition $\vec u\perp \vec a$: $\vec u\cdot \vec a=0.$ Now, $\vec a\cdot \vec a=2^2+3^2+(-1)^2=14,$ $\vec b\cdot \vec a=0\cdot2+1\cdot3+1\cdot(-1)=2.$ Hence, $\vec u\cdot \vec a=(x\vec a+y\vec b)\\cdot \vec a=x(\vec a\cdot \vec a)+y(\vec b\cdot \vec a)=14x+2y=0.$ So, $7x+y=0\quad\Rightarrow\quad y=-7x.$

3. Use the condition $\vec u\cdot \vec b=24$: $\vec a\cdot \vec b=2,\qquad \vec b\cdot \vec b=0^2+1^2+1^2=2.$ Therefore, $\vec u\cdot \vec b=(x\vec a+y\vec b)\cdot \vec b=x(\vec a\cdot \vec b)+y(\vec b\cdot \vec b)=2x+2y=24.$ So, $x+y=12.$ Substitute $y=-7x$: $x-7x=12\Rightarrow -6x=12\Rightarrow x=-2.$ Thus, $y=14.$

4. Find $\vec u$: $\vec u=-2\vec a+14\vec b.$ Compute it: $-2\vec a=(-4,-6,2),\qquad 14\vec b=(0,14,14).$ Hence, $\vec u=(-4,8,16).$

5. Now calculate $|\vec u|^2$: $|\vec u|^2=(-4)^2+8^2+16^2=16+64+256=336.$

6. Compare with options:

• A: $336$ ✅
• B: $315$
• C: $256$
• D: $84$

Therefore, the correct answer is $\boxed{336}.$