A*B*C = A*C*B
It's not operator location that implies precedence  it's only operator type and parens that set that.
Probably the only reason the confusion ever existed is because at some point around 20 years ago compilers popularly ended up whenever there was code like this and evaluating multiplications first, then divisions  leading to computation tools that enforced the thinking that leads to the wrong conclusion.
6

2(1+2)
and
6(1+2)

2
are both correctly handwritten notations of two different problems.
6/2(1+2) is machineform (the typewriter) notation  and so only the conservative and strict rules may apply, to free from compiler confusions.
A * B * C = A * C * B
A / B * C = A * C / B
Therefore = 9.
