In a non-right-angled triangle PQR, let p, q, r denote the lengths...

By sine rule in $PQR$ $\dfrac{p}{sinP} = \dfrac{q}{sinQ} = \dfrac{r}{sinR} = 2R$ $\dfrac{\sqrt{3}}{sinP} = \dfrac{1}{sinQ} = \dfrac{1}{sinR} = 2$\left(1\right)$sinP = \dfrac{\sqrt{3}}{2}$ and $sinQ = sinR = \dfrac{1}{2}$ $P = {60}^{°}$ or ${120}^{°}, Q = {30}^{°}$ and ${150}^{°}$ $⇒$ As $PQR$ is not a right angle triangle Only possibility is $P = {120}^{°}, Q = R = {30}^{°}$ $⇒ PQR$ is an isosceles triangle, hence, $PE$ is also a median and $‘O’$ will be centroid. ( $∆ PQE$ and $∆ PRE$ are congruent and $E$ is mid point of $QR$ ) $⇒ r = q = 1, P = \sqrt{3}$ Option $\left(A\right):$ Length of median from $R$ $RS = \dfrac{1}{2}\sqrt{2{p}^{2} + 2{q}^{2} - {r}^{2}} = \dfrac{1}{2}\sqrt{2{\left(\sqrt{3}\right)}^{2} + 2 ( 1 ) - 1}$ = \dfrac{\sqrt{7}}{2}$Option$\left(B\right):$Now, area of$\delta SEF = \dfrac{1}{4}\delta PQR$...$\left(i\right)$and area of$\delta SOE = \dfrac{1}{3}\delta SEF$...$\left(ii\right)$from$\left(i\right)$and \left(ii\right)$\delta SOE = \dfrac{1}{12}\delta PQR$...$\left(iii\right)$\delta PQR = \dfrac{1}{2}pqsinR = \dfrac{1}{2} \times \sqrt{3} \times 1 \times \dfrac{1}{2} = \dfrac{\sqrt{3}}{4}$ $\delta SOE = \dfrac{1}{12}\delta PQR = \dfrac{\sqrt{3}}{48}$ Option $\left(D\right):$ Length of $OE$ As $O$ is centroid $OE$ will be $\dfrac{1}{3}$ of $PE$ $PE = \dfrac{1}{2}\sqrt{2{q}^{2} + 2{r}^{2} - {p}^{2}} = \dfrac{1}{2}\sqrt{2 + 2 - 3}$ $= \dfrac{1}{2}$ $OE = \dfrac{1}{3}PE = \dfrac{1}{6}$ Option $\left(C\right):$ Radius of incircle $= \dfrac{\Delta}{S}$\left(i . e . \dfrac{area}{semi - perimeter}\right)$= \dfrac{\dfrac{\sqrt{3}}{4}}{\left(\dfrac{1 + 1 + \sqrt{3}}{2}\right)}$ $= \dfrac{\sqrt{3}}{2\left(2 + \sqrt{3}\right)}$= \dfrac{\sqrt{3}}{2}\left(2 - \sqrt{3}\right)