solubility product $\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{s})^{2} \times \mathrm{S}=4 \mathrm{s}^{3}$

$\mathrm{K}_{\mathrm{sp}}=\left(1.1 \times 10^{-12}\right)$

$s=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=0.65 \times 10^{-4}$

$\mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$

$\mathrm{K}_{\mathrm{sp}}=5 \times 5 \quad\left(\mathrm{K}_{\mathrm{sp}}=1.8 \times 10^{-10}\right)$

$s=\sqrt{\mathrm{K}_{\mathrm{sp}}}=1.34 \times 10^{-5}$

AgBr $\rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$

$\mathrm{K}_{\mathrm{sp}}=5 \times 5 \quad\left(\mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13}\right.$

$S=\sqrt{K_{s p}}=0.71 \times 10^{-6}$

$\mathrm{Agl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-}$

$\left.\mathrm{Ksp}=\mathrm{S} \times \mathrm{S} \quad \mathrm{K}_{\mathrm{sp}}=8.3 \times 10^{-17}\right)$

$S=\sqrt{K_{\mathrm{sp}}}=0.9 \times 10^{-8}$

therefore, solubility of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is highest so it will precipitate last.