I don't know if you were looking for some number-theoretic insights (or even whether any exist to be found), but a brute-force computer program can easily find all such pairs of triples:
(A, B, C) (X, Y, Z) (Sum, Product)
(8, 12, 15) (9, 10, 16) (35, 1440)
(3, 8, 10) (4, 5, 12) (21, 240)
(5, 9, 14) (6, 7, 15) (28, 630)
(4, 9, 10) (5, 6, 12) (23, 360)
(3, 10, 12) (4, 6, 15) (25, 360)
(4, 10, 14) (5, 7, 16) (28, 560)
(6, 10, 14) (7, 8, 15) (30, 840)
(4, 8, 15) (5, 6, 16) (27, 480)
(6, 12, 14) (7, 9, 16) (32, 1008)
("All" up to swapping (A,B,C) and (X,Y,Z), of course.)
Python code if anyone's interested:
ss = {} #Triples which give a certain (sum, product)
for A in range(3,18):
for B in range(A+1, 18):
for C in range(B+1, 18):
p = (A+B+C, A*B*C)
ss[p] = ss.get(p, []) + [(A,B,C)]
for p in ss:
if len(ss[p])>=2:
print ss[p], "\t", p
As for solving it manually, I don't think there is any method that is significantly different from brute force. One can prune the list of choices to consider, but it will still take exhaustive enumeration or trial-and-error to find such triples. An ad hoc method for an ad hoc problem. :-)
For instance — going by trial-and-error and blind guesswork — I might start by trying (4,5) for (A,B). Then 20C = XYZ suggests maybe trying X=10 (because all prime factors of 20 must occur somewhere on the right), after which the equations become {C=1+Y+Z, 2C=YZ}, and you know one of Y,Z must be even; Y=6 doesn't work and Y=8 happens to give a valid solution Z=3. This (after you order them correctly) is one valid pair of triples. But other guesses may lead to lots of blind alleys and backtracking, so I don't really recommend this method. Then again, I suspect there is nothing much better.