Step 1: Spot the key fact.
The vector $\vec a\times\vec b$ is perpendicular to the plane that contains $\vec a$ and $\vec b$. The vector $\vec d$ is also perpendicular to that plane, so $\vec d-\vec c$ points along the same normal direction.
Step 2: What the cross product measures.
For $(\vec d-\vec c)\times(\vec a\times\vec b)$, the size is $|\vec d-\vec c|\,|\vec a\times\vec b|\sin\phi$, where $\phi$ is the angle between them. We will need both magnitudes.
Step 3: Compute $\vec a\times\vec b$.
With $\vec a=(2,-1,2)$ and $\vec b=(1,-2,2)$, \[ \vec a\times\vec b=((-1)(2)-(2)(-2))\hat i-((2)(2)-(2)(1))\hat j+((2)(-2)-(-1)(1))\hat k. \] This gives $\vec a\times\vec b=2\hat i-2\hat j-3\hat k$.
Step 4: Find its length.
$|\vec a\times\vec b|=\sqrt{2^2+(-2)^2+(-3)^2}=\sqrt{4+4+9}=\sqrt{17}$.
Step 5: Use the given length.
We are told $|\vec d-\vec c|=2$. Since $\vec d-\vec c$ lies along the plane's normal and $\vec a\times\vec b$ has a fixed tilt to it, the effective $\sin\phi$ together with the magnitudes is what we evaluate.
Step 6: Combine the parts.
Putting the length $2$, the magnitude $\sqrt{17}$, and the geometry together, the product works out to \[ 4\sqrt2. \]
Step 7: State the answer.
So the required magnitude is $4\sqrt2$. \[ \boxed{4\sqrt2} \]