1. Let $\vec a=(2,-1,2),\qquad \vec b=(1,2,-1).$ Since $\vec v$ lies in the plane containing $\vec a$ and $\vec b$, write $\vec v=x\vec a+y\vec b.$ So, $\vec v=x(2,-1,2)+y(1,2,-1)=(2x+y,-x+2y,2x-y).$

2. Given that $\vec v$ is perpendicular to $(3,2,-1)$, so $\vec v\cdot (3,2,-1)=0.$ Substitute $\vec v$: $3(2x+y)+2(-x+2y)-(2x-y)=0.$ Simplify: $6x+3y-2x+4y-2x+y=0$ $2x+8y=0$ $x+4y=0\implies x=-4y.$

3. The projection of $\vec v$ on $\vec a$ is $19$ units. Scalar projection is $\frac{\vec v\cdot \vec a}{|\vec a|}=19.$ Now, $|\vec a|=\sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3.$ Hence, $\vec v\cdot \vec a=57.$

Now compute $\vec v\cdot \vec a$ using $\vec v=x\vec a+y\vec b$: $\vec v\cdot \vec a=x(\vec a\cdot \vec a)+y(\vec b\cdot \vec a).$ We have $\vec a\cdot \vec a=9,$ $\vec a\cdot \vec b=2(1)+(-1)(2)+2(-1)=2-2-2=-2.$ Thus, $9x-2y=57.$ Using $x=-4y$, $9(-4y)-2y=57$ $-36y-2y=57$ $-38y=57$ $y=-\frac{3}{2},\qquad x=6.$

4. Therefore, $\vec v=6\vec a-\frac{3}{2}\vec b.$ Compute coordinates: $\vec v=6(2,-1,2)-\frac{3}{2}(1,2,-1)$ $=(12,-6,12)-\left(\frac32,3,-\frac32\right)$ $=\left(\frac{21}{2},-9,\frac{27}{2}\right).$

5. Now find $|\vec v|^2$: $|\vec v|^2=\left(\frac{21}{2}\right)^2+(-9)^2+\left(\frac{27}{2}\right)^2$ $=\frac{441}{4}+81+\frac{729}{4}$ $=\frac{1170}{4}+81$ $=\frac{585}{2}+81$ $=\frac{747}{2}.$

6. We need $|2\vec v|^2=4|\vec v|^2=4\cdot \frac{747}{2}=1494.$

Therefore, the required integer is $\boxed{1494}.$