Plz help with this question

Dear student

T h e e l e c t r i c f i e l d d u e t o a p o i n t c h a r g e i s g i v e n b y \vec{E} = \frac{1}{4 π ε_{o}} \frac{q}{r^{2}} \hat{r} w h e r e \hat{r} i s t h e d i r e c t i o n o f t h e e l e c t r i c f i e l d . S o e l e c t r i c f i e l d a t p o i n t A \vec{E} = \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.05)}^{2}} \hat{x} + \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.05)}^{2}} (\hat{x}) \vec{E} = \frac{1}{4 π ε_{o}} \frac{2 \times 10^{- 8}}{{(0.05)}^{2}} \hat{x} \vec{E} = 9 \times 10^{9} \times \frac{2 \times 10^{- 8}}{{(0.05)}^{2}} \hat{x} \vec{E} = 7.2 \times 10^{4} \hat{x} N / C A t p o i n t B \vec{E} = \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.05)}^{2}} (- \hat{x}) + \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.15)}^{2}} (\hat{x}) \vec{E} = \frac{1}{4 π ε_{o}} [\frac{10^{- 8}}{{(0.15)}^{2}} - \frac{10^{- 8}}{{(0.05)}^{2}}] \hat{x} \vec{E} = 9 \times 10^{9} \times 10^{- 8} \times 355.55 (- \hat{x}) \vec{E} = 3.2 \times 10^{4} (- \hat{x}) N / C A t p o i n t C \vec{E} = \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.01)}^{2}} (\cos 60^{°} \hat{x} + \sin 60^{°} \hat{y}) + \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.01)}^{2}} (\cos 60^{°} \hat{x} - \sin 60^{°} \hat{y}) \vec{E} = \frac{1}{4 π ε_{o}} \frac{10^{- 8}}{{(0.01)}^{2}} \times 2 \cos 60^{°} \hat{x} \vec{E} = 9 \times 10^{9} \times \frac{10^{- 8}}{{(0.01)}^{2}} \times 2 \times \frac{1}{2} \hat{x} \vec{E} = 9 \times 10^{5} \hat{x} N / C

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