What's wrong in my solution? Ways of choosing 5 items from 3 catagory with 3, 6, 14 items, while having 1 item from each catagory.

You are overcounting many, many choices, such as the following selections:

• train app A, SNS app B, games app C, games app D, games app E
• train app A, SNS app B, games app D, games app C, games app E

Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.

As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.

My attempt(casework) is the following:

The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.

Case 1: 1, 1, 3

There are different combinations based on the different choices of the largest group. So we compute them separately, giving

$$\binom{3}{1}\binom{6}{1}\binom{14}{3} + \binom{3}{1}\binom{6}{3}\binom{14}{1} + \binom{3}{3}\binom{6}{1}\binom{14}{1} = 7476.$$

Case 2: 1, 2, 2

Similarly the the last one, we have $$\binom{3}{1}\binom{6}{2}\binom{14}{2}+\binom{3}{2}\binom{6}{1}\binom{14}{2}+\binom{3}{2}\binom{6}{2}\binom{14}{1} = 6363.$$

Summing the two cases gives a total of $$\boxed{13839}.$$

As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.

So, how to avoid this overcounting?

Take 3T,6S and 14G apps.
In total, you want 5 apps.

So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$ Now, count for number of ways you have for each of the above selection, and add.

$$(^3C_1^6C_1^{14}C_3) + (^3C_1^6C_3^{14}C_1) + (^3C_3^6C_1^{14}C_1) + (^3C_2^6C_1^{14}C_1) + (^3C_2^6C_1^{14}C_2) + (^3C_1^6C_2^{14}C_2) = 13839$$