The least number which leaves remainder 2 when divided by 5, 6 or 8, and which is exactly divisible by 7, is:

Numbers leaving remainder 2 for all three divisors have the form $120k+2$, because LCM(5, 6, 8) = 120. Testing for divisibility by 7: 122, 242, 362 and 482 all fail, but $602=7\times 86$ works. Check: $602=120\times 5+2$, so it also leaves remainder 2 on division by 5, 6 and 8.