For shown situation, what will be the magnitude of minimum force in newton that can be applied in any direction so that the resultant force is along east direction?

Let force be F so resultant is in east direction 

$\begin{array}{l} 4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\left(5 \cos {37}^{\circ }\hat{\mathrm{i}}+5 \sin {37}^{\circ }\hat{\mathrm{j}}\right)+\overrightarrow{\mathrm{F}}=\mathrm{k}\hat{\mathrm{i}}\\ \Rightarrow 4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\overrightarrow{\mathrm{F}}=\mathrm{k}\hat{\mathrm{i}}\Rightarrow 8\hat{\mathrm{i}}+6\hat{\mathrm{j}}+\overrightarrow{\mathrm{F}}=\mathrm{k}\hat{\mathrm{i}}\\ \Rightarrow \overrightarrow{\mathrm{F}}=(\mathrm{k}-8)\hat{\mathrm{i}}-6\hat{\mathrm{j}}\Rightarrow \mathrm{F}=\sqrt{(\mathrm{k}-8{)}^2+(6{)}^2} \\ \Rightarrow {\mathrm{F}}_{min}=6 \mathrm{N}\end{array}$