Let magnitude be equal to $$\lambda$$

$$\overrightarrow {OA} = \lambda \left[ {\cos 30^\circ \widehat i + \sin 30\widehat j} \right] = \lambda \left[ {{{\sqrt 3 } \over 2}\widehat i + {1 \over 2}\widehat j} \right]$$

$$\overrightarrow {OB} = \lambda \left[ {\cos 60^\circ \widehat i - \sin 60\widehat j} \right] = \lambda \left[ {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right]$$

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

$$\therefore$$ $$\overrightarrow {OA} + \overrightarrow {OB} - \overrightarrow {OC} $$

$$ = \lambda \left[ {\left( {{{\sqrt 3 + 1} \over 2} + {1 \over {\sqrt 2 }}} \right)\widehat i + \left( {{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \right)\widehat j} \right]$$

$$\therefore$$ Angle with x-axis

$${\tan ^{ - 1}}\left[ {{{{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \over {{{\sqrt 3 } \over 2} + {1 \over 2} + {1 \over {\sqrt 2 }}}}} \right] = {\tan ^{ - 1}}\left[ {{{\sqrt 2 - \sqrt 6 - 2} \over {\sqrt 6 + \sqrt 2 + 2}}} \right]$$

$$ = {\tan ^{ - 1}}\left[ {{{1 - \sqrt 3 - \sqrt 2 } \over {\sqrt 3 + 1 + \sqrt 2 }}} \right]$$

Hence option (a).