1. Given vectors $\vec a = 2\hat i + \hat j - \hat k = (2,1,-1), \qquad \vec b = \hat i + 2\hat j + \hat k = (1,2,1).$

2. Form of $\vec c$ $\vec c = \alpha \vec a + \beta \vec b.$ Since $\vec c$ is a linear combination of $\vec a,\vec b$, it lies in the plane spanned by $\vec a,\vec b$.

3. Condition on projection of $\vec c$ on $(\vec a+\vec b)$

   First compute: $\vec a + \vec b = (3,3,0).$ Its magnitude is $|\vec a+\vec b| = \sqrt{3^2+3^2} = 3\sqrt2.$

   The scalar projection of $\vec c$ on $(\vec a+\vec b)$ is given as $3\sqrt2$: $\frac{\vec c\cdot (\vec a+\vec b)}{|\vec a+\vec b|} = 3\sqrt2.$ Since $|\vec a+\vec b|=3\sqrt2$, we get $\vec c\cdot (\vec a+\vec b) = (3\sqrt2)(3\sqrt2)=18.$

4. Expression to minimize We need minimum of $(\vec c-(\vec a\times \vec b))\cdot \vec c.$ Expand: $(\vec c-(\vec a\times \vec b))\cdot \vec c = \vec c\cdot \vec c - (\vec a\times \vec b)\cdot \vec c.$

   But $\vec a\times \vec b$ is perpendicular to both $\vec a$ and $\vec b$, hence perpendicular to every linear combination of them, including $\vec c$. Therefore, $(\vec a\times \vec b)\cdot \vec c = 0.$ So the required quantity reduces to $\vec c\cdot \vec c = |\vec c|^2.$

   Thus we must minimize $|\vec c|^2$ subject to $\vec c\cdot (\vec a+\vec b)=18.$

5. Minimum of $|\vec c|^2$ under the projection constraint

   Among all vectors with fixed dot product with a given vector $\vec d$, the minimum magnitude occurs when the vector is parallel to $\vec d$.

   Here $\vec d=\vec a+\vec b=(3,3,0)$, and this vector itself lies in the span of $\vec a,\vec b$, so it is an allowed choice.

   Let $\vec c = \lambda (\vec a+\vec b).$ Then $\vec c\cdot (\vec a+\vec b)=\lambda |\vec a+\vec b|^2 = \lambda \cdot 18.$ Given this equals $18$, we get $\lambda=1.$ Hence the minimizing vector is $\vec c = \vec a+\vec b.$

6. Minimum value Therefore, $|\vec c|^2 = |\vec a+\vec b|^2 = 18.$

So the minimum value of $(\vec c-(\vec a\times \vec b))\cdot \vec c$ is $\boxed{18}.$