Answer: 925

Small Hint:

The three representations give $196a + 14b + c = 225a + 15c + b$ $= 222a + 37c$

Big Hint:

Comparing the last two gives $b = 22c - 3a;$ substituting into the first forces $a = 4c,$ where $a$ and $c$ are base-$6$ digits

Solution:

Writing out the place values, $n = 196a + 14b + c$ $= 225a + 15c + b$ $= 222a + 37c,$ where $a$ and $c$ are base-$6$ digits with $1 \le a \le 5$ and $0 \le c \le 5,$ and $0 \le b \le 13.$

Equating the last two expressions gives $b = 22c - 3a.$ Substituting into $196a + 14b + c = 225a + 15c + b$ (which says $13b = 29a + 14c$) yields $13(22c - 3a) = 29a + 14c,$ so $272c = 68a,$ that is $a = 4c.$ The digit bounds force $c = 1,$ $a = 4,$ and then $b = 22 - 12 = 10,$ which is a valid digit in bases $14$ and $15.$

Therefore $n = 222 \cdot 4 + 37 \cdot 1 = 925.$ Indeed $925 = 196 \cdot 4 + 14 \cdot 10 + 1,$ confirming the base-$14$ form. The answer is $925.$