Total ways 52C3 = 22100
There are 4 suit in a pack of cards so three suit can be selected in 4C3 ways.
One card each from different unit can selected as = 13C1 X 13C1 X 13C1 ways
So, favourable ways = 4C3 X 13C1 X 13C1 X 13C1 = 8788
∴ Required probability = 8788/22100 = 169/425