Given a, b, c are positive integers forming an increasing G.P., b – a is a prefect square of a natural number, and log₆a+log₆b + log₆c = 6. Find the value of (a + b + c – 105)

a, b, c are in G.P. b² = ac
log₆a + log₆b + log₆c = 6
a b c = 6⁶ b³ = 6⁶ b = 6² = 36
$$\begin{gathered} \mathrm{ac}=36\times 36=2^4\times 3^4 \\ \mathrm{b}-\mathrm{a}={\mathrm{N}}^2 \\ 36-\mathrm{a}={\mathrm{N}}^2 \end{gathered}$$
a is factor of 2⁴ 3⁴ a = 27 is possible value
36 – 27 = 9 = (3)² $\Rightarrow$ a = 27,
b = 36, c = 48

a + b + c = 111