$\underset{1\text{mol}}{4\mathrm{A}}+\underset{0.6\text{mol}}{2\mathrm{B}}+\underset{0.72\text{mol}}{3\mathrm{C}}\to {\mathrm{A}}_4{\mathrm{B}}_2{\mathrm{C}}_3$

$\begin{array}{ccc}\frac{1}{4};& \frac{0.6}{2};& \frac{0.72}{3}\\ 0.25;& 0.3;& 0.24\end{array}$
In the present case, reactant 'C' will be the limiting reactant because it will give least amount of product on being completely consumed.
As 3 mol C gives 1 mol product,
So 0.72 mol C will give 0.24 mol of product.