The magnitude of vectors \(\overrightarrow{ OA }, \overrightarrow{ OB }\) and \(\overrightarrow{ OC }\) in the given figure are equal. The direction of \(\overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }\) with \(x\)-axis will be

\(\overrightarrow{ OA }=\lambda\left[\cos 30^{\circ} \hat{ i }+\sin 30 \hat{ j }\right]=\lambda\left[\frac{\sqrt{3}}{2} \hat{ i }+\frac{1}{2} \hat{ j }\right]\)

\(\overrightarrow{ OB }=\lambda\left[\cos 60^{\circ} \hat{ i }-\sin 60 \hat{ j }\right]=\lambda\left[\frac{1}{2} \hat{ i }-\frac{\sqrt{3}}{2} \hat{ j }\right]\)

\(\overrightarrow{ OC }=\lambda\left[\cos 45^{\circ}(-\hat{ i })+\sin 45 \hat{ j }\right]=\lambda\left[-\frac{1}{\sqrt{2}} \hat{ i }+\frac{1}{\sqrt{2}} \hat{ j }\right]\)

\(\therefore \overrightarrow{ OA }+\overrightarrow{ OB }-\overrightarrow{ OC }\)

\(=\lambda\left[\left(\frac{\sqrt{3}+1}{2}+\frac{1}{\sqrt{2}}\right) \hat{ i }+\left(\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\right) \hat{ j }\right]\)

\(\therefore\) Angle with \(x\)-axis

\(\tan ^{-1}\left[\frac{\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt{2}}}\right]=\tan ^{-1}\left[\frac{\sqrt{2}-\sqrt{6}-2}{\sqrt{6}+\sqrt{2}+2}\right]\)

\(=\tan ^{-1}\left[\frac{1-\sqrt{3}-\sqrt{2}}{\sqrt{3}+1+\sqrt{2}}\right]\)