Help Please?

$BBA_6+41B_6+A15_6=A152_6 \\ \text{looking at the 1's digit we have, (all numbers will be base 6)} \\ A+B+5 =2 \mod 6 \\ \text{given the possible ranges of A and B we have}\\ A+B = 9 \text{ with a carry of 2 or }\\ A+B = 3 \text{ with a carry of 1}$

$\text{let's suppose the first case}\\ \text{then looking at the 6's digit we have}\\ B+1+1+2 = 5 \mod 6\\ B+4 = 5 \mod 6 \\ B = 7 \text{ or }B=1 \\ \text{well B=7 is clearly out as B<6, and A+B=9, which would make A=8, which is also out}$

$\text{so let's assume the 2nd case}\\ \text{looking at the 6's digit we have}\\ B+1+1+1 = 5 \mod 6\\ B+3 = 5 \mod 6\\ B=8 \text{ or }B=2 \\ \text{again B=8 is clearly out. If B=2 then A=1 and this seems like a solution}$

$\text{checking we have }\\ 221 + 412+115 = 1152 \text{ all in base 6}$

$|A-B| = |1-2|=1$

We have the following equation

( 36B + 6B + A )  + ( 4*36 + 1* 6 + B)  + (36A  + 6  + 5)  = 216A + 36 + 5*6 + 2

Simplify

43B  + 37A  + 150 +  11  = 216A + 68

43B + 37A  + 161  = 216A + 68

43B   = 179A - 93

Note....this equation will be true when B  = 2  and A  = 1

So

l A  - B  l  =   l 1   - 2   l    =  l  -1  l   =   1