We need to check from the options that which of them must be true.

(A) $\(\angle a=\angle b+\angle c\)$, if line $Q R$ is parallel to line NO As $\(\mathrm{RQ} \| \mathrm{NO}\)

$, we get angle $\(\mathrm{NOM}=\)$ angle PRQ (= Corresponding angles) i.e. angle $\(\mathrm{PRQ}=\mathrm{c}\)$

Now, angle a is an exterior angle of triangle PQR , so we get $\(\mathrm{a}=\)$ angle $\(\mathrm{PQR}+\)$ angle $\(\mathrm{PRQ}=\mathrm{b}+\mathrm{c}\)$ (Exterior angle of a triangle is equal to the sum of the interior opposite angles)

So, option (A) is true.

(B) $\(\angle a+\angle M=180\)$, if line $P Q$ is parallel to line $\(M N-\)$ If PQ is parallel to line MN , we get $\(\angle \mathrm{QPM}+\angle \mathrm{NMP}=180^{\circ}\)$ (the sum of the interior angles of one side of transversal is supplementary) i.e. $\left(\(180^{\circ}-\mathrm{a}\right)+\left(180^{\circ}-\angle \mathrm{NMO}\right)=180^{\circ} \Rightarrow \mathrm{a}+\angle \mathrm{NMO}=180^{\circ}=\mathrm{a}+\angle \mathrm{M}\)$. So, option (B) is true.

(C) $\(\angle N=b\)$, if the above two options A and B are true besides being given that $O M N$ is an equilateral triangle - If RQ\|NO\& $\(P Q \| M N\)$, we get $\(\triangle P Q R \sim \triangle M N O\)$ ( $\(\because \angle \mathrm{QPR}=\angle \mathrm{NMO} \& \angle \mathrm{QRP}=\angle \mathrm{NOM}\)$ Corresponding angles). So, the third angle of the triangle of the triangles must also be the same i.e. $\(\angle \mathrm{N}=\angle \mathrm{PQR}=\mathrm{b}\)$. Thus, statement (C) is true.

Hence all three options (A), (B) \& (C) are true.