1. Let the common magnitude of the mutually perpendicular vectors $\vec a, \vec b, \vec c$ be $m$.

   So, $|\vec a|=|\vec b|=|\vec c|=m$ and $\vec a\cdot \vec b=\vec b\cdot \vec c=\vec c\cdot \vec a=0.$

2. The vector with which they are equally inclined is $\vec a+\vec b+\vec c.$ Let $\vec s=\vec a+\vec b+\vec c.$

3. Since each of $\vec a,\vec b,\vec c$ makes the same angle $\theta$ with $\vec s$, use the dot product formula: $\cos\theta=\frac{\vec a\cdot \vec s}{|\vec a|\,|\vec s|}.$

4. Compute $\vec a\cdot \vec s$: $\vec a\cdot \vec s=\vec a\cdot(\vec a+\vec b+\vec c)=|\vec a|^2+\vec a\cdot\vec b+\vec a\cdot\vec c=m^2.$

5. Compute $|\vec s|$: $|\vec s|^2=(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)$ $=|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$ $=m^2+m^2+m^2=3m^2.$ Hence, $|\vec s|=m\sqrt{3}.$

6. Therefore, $\cos\theta=\frac{m^2}{m\cdot m\sqrt{3}}=\frac{1}{\sqrt{3}}.$

7. Now find $\cos 2\theta$: $\cos 2\theta=2\cos^2\theta-1=2\left(\frac{1}{3}\right)-1=-\frac{1}{3}.$

8. Then $36\cos^2 2\theta=36\left(\frac{1}{9}\right)=4.$

So the required integer is $\boxed{4}.$