$4\mathrm{A}+2\mathrm{B}+3\mathrm{C}\to {\mathrm{A}}_4{\mathrm{B}}_2{\mathrm{C}}_3$
$1$ $\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$ $0.6$ $\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$ $\mathrm{}0.72$ $\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$
In the above reaction, reactant $\mathrm{\text{'}}\mathrm{C}\mathrm{\text{'}}\mathrm{}$ will be the limiting reactant and it will decide yield product.
$\because$ from $3$ moles of $\mathrm{C}$ ; $1$ mole product is formed
$\therefore$ from $0.72$ moles of $\mathrm{C}=\frac{1}{3}\times 0.72$
$=0.24$ moles of product