The function |2-3x| is continuos ∀x∊R except at x=2/3 as if x>2/3 (i.e in RHL case) then function will become f(x)= 2-3x
if x<2/3(i.e in LHL case) then function will become f(x)=3x-2. Now, if you put any value of x∈R in these equation you will always get some value i.e function is continuos for LHL and RHL both.
For differentiability, for LHL(Left Hand Limit) i.e. if x<2/3 then value of function at which min/max occur will be 3
and for RHL(Right Hand Limit) for x>2/3 the value of function at which min/max occur will be -3.
Since both are different value hence, function cannot be differentiable at x=2/3
Hence, option C is the answer.