I know of one with 12 vertices, 20 faces, and 30 edges. Are there any others? Almost sorry I asked the question. The answer is trivial. Use Euler's formula and the unique solution pops out.
Let $v,e,f$ be the number of vertices, edges and faces. So by your given condition, we have $ e = 5v/2 , f = 5v/3 $. And use this in the euler characteristic formula for sphere $ v-e+f =2$ to get $v (1-5/2+5/3) = 2 \implies v = 12 $. So there are only one such triangulation.