Consider a uniform spherical charge distribution of radius $R_1$   centered at the origin O. In this distribution, a spherical cavity of radius  $R_2$, centered at P with distance $OP=a=R_1-R_2\left(\text{see\hspace{0.17em}figure}\right)$  is made . If the electric field inside the cavity at position  $\overrightarrow{r}$  is $\overrightarrow{E}\stackrel{\rightharpoonup }{\left(r\right)}$ , then the correct statement(s) is (are) 

Let us consider a point M inside the cavity where electric field has to be calculated . Assume the cavity to contain similar charge distribution of positive and negative charge as the rest of sphere electicfiled at M due to uniformly distributed charge of the whole sphere of radius $R_1 {\overrightarrow{E}}_1=\frac{\rho }{3\epsilon }\overrightarrow{r}$   
Electric field at M due to the negative charge distribution in the cavity   

${\overrightarrow{E}}_2=\frac{\rho }{3\epsilon }\stackrel{\rightharpoonup }{MP}$
  The total electric field at M inside  the cavity,                               
   $$\begin{gathered} \overrightarrow{E}=\overrightarrow{E}1+{\overrightarrow{E}}_2=\frac{\rho }{3\epsilon }\overrightarrow{r}+\frac{\rho }{3\epsilon }\stackrel{\rightharpoonup }{MP} \\ \therefore \overrightarrow{E}=\frac{\rho }{3\epsilon }\overrightarrow{r}+\frac{\rho }{3\epsilon }(\overrightarrow{a}-\overrightarrow{r})[\therefore \overrightarrow{r}+\stackrel{\rightharpoonup }{MP}=\overrightarrow{a}] \\ \therefore \overrightarrow{E}=\frac{\rho }{3\epsilon }\overrightarrow{a} \end{gathered}$$