Answer:
Option A,B
Explanation:
PQ= (2) R sin 60°

= $(2R)\frac{\sqrt{3}}{2}=(\sqrt{3}R)$
$q_{enclosed}=\lambda(\sqrt{3}R)$
We have, $\phi =\frac{q_{enclosed}}{\epsilon_{0}}$
$\Rightarrow$ $\phi =(\frac{\sqrt{3}\lambda R_{}}{\epsilon_{0}})$
Also, electric field is perpendicular to wire, so Z- component will be zero.