$A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$
Taking C common,
=$(A\bar{B} + \bar{A}B + \bar{A}\bar{B} ) . C$
=$((A \bigoplus B) + \bar{A}\bar{B} ) . C$
Remember this shortcut, [ $(P \bigoplus Q) + PQ => P + Q $], also we know that $A \bigoplus B = \bar{A} \bigoplus \bar{B}$, negating both inputs have no effect on output of XOR and XNOR. Hence we can write
$(A \bigoplus B) + \bar{A}\bar{B} = \bar{A}+\bar{B} $
Hence it directly becomes,
=$(\bar{A}+\bar{B}).C$
=$\bar{A}.C+\bar{B}.C$
Another method is to draw K-MAP from given min terms and then minimize :
$f(A,B,C) = A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$
101 011 001
$f(A,B,C) = \sum \small m (1,3,5)$