Space between four spheres

First let's calculate the volume of the tetrahedron. Its base is an equilateral triangle of area $A=\sqrt{3}$ and its height is $h=\frac{2}{3}\sqrt{6}$

So its volume is $V_{tetr}=\frac{1}{3}Ah = \frac{2}{3}\sqrt{2}$

Next we determine the solid angle of a tetrahedron (as seen from one of its vertices) From, for example, in Wikipedia , we use the formula. $\tan\frac{Ω}{2}= \frac{|\vec{a}\cdot(\vec{b}×\vec{c})|} {|a||b||c|+|c|\vec{a}\cdot\vec{b}+|a|\vec{b}\cdot\vec{c}+|b|\vec{c}\cdot\vec{a}}$

Using appropriate vectors that correspond to the edges from one vertex, for example $\vec{a}=(2,0,0), \vec{b}=(1,\sqrt{3},0), \vec{ c}= (1, \frac{1}{3}\sqrt{3}, \frac{2}{3}\sqrt{6})$ , we get $\tan\frac{Ω}{2} = \frac{ 4\sqrt{2}} {8+4+4+4} =\frac{ \sqrt{2}} {5}$

Drawing a right-angled triangle, with right sides $5$ and $\sqrt{2}$ , we get a hypothenuse of $3\sqrt{3}$ and  see that $\cos \frac{Ω}{2} = \frac{5}{ 3\sqrt{3}}, \sin \frac{Ω}{2} = \frac{\sqrt{2}}{ 3\sqrt{3}}$ Now using the double angle formula we find $\cos Ω = \frac{25}{27} - \frac{2}{27} = \frac{23}{27}$ and hence $Ω=\arccos{\frac{23}{27}}$ The volume of a sphere of radius 1 equals $\frac{4}{3}π$ , and the solid angle for a full sphere is 4π. So, the volume of the intersction of a sphere with the tetrahedron is $V_{intersect} = \frac{Ω}{3}$ . Now we have 4 such intersections, the volume between the spheres is $V_{tetr} -4 V_{intersect} = \frac{2}{3}\sqrt{2}-\frac{4Ω}{3} = \frac{2\sqrt{2}-4\arccos{\frac{23}{27}}}{3}$ So the answer is $2+2+4+23+27+3=\boxed{61}$