36, 64, and 81 are all perfect squares. That means they can all be written as $ 1\cdot a^2+0\cdot a^1+0\cdot a^0=100_a $, where $ a $ is the square root of each number. So, all three numbers can be represented by the digits $ \boxed{100} $ when converted to other bases. To see that no other digits work, note that only the bases 4, 5, and 6 use three digits to represent the number 36. (This follows from $ b^2\leq 36<b^3 $, which expresses the condition that 36 has 3 digits in base $ b $). The representations are $ 100_6 $, $ 121_5 $ and $ 210_4 $, only one of which fits the form $ \triangle\Box\Box $.