It looks like there are two equations with three unknowns for which there are many solutions. So solve() gives the results for two of the variables in terms of the third, which you can set to an arbitray value. In this case, it chose to solve for a1 and a2 in terms of a3. But you can make it solve for any two in terms of the third, for example a2 and a3 in terms of a1
syms a3 a2 a1
eqn = [a3-3*a2+10*a1==68,5*a3+6*a2+2*a1==31];
S = solve(eqn,[a2 a3])
Now pick an arbitrary value for a1, say sqrt(3), and show that the solution satisfies the eqn
subs(eqn,[a1 a2 a3],[sqrt(3) subs([S.a2 S.a3],a1,sqrt(3))])
