In the given figure, if OQ = QR, then the value of m is:

In $$\triangle\ $$OQR,

OQ = QR

$$=$$>  $$\angle\ $$ORQ = $$\angle\ $$QOR

$$=$$>  $$n^{\circ\ }$$ = $$\angle\ $$QOR

$$=$$>  $$\angle\ $$QOR = n$$^{\circ\ }$$

In $$\triangle\ $$OQR,

$$\angle\ $$ORQ + $$\angle\ $$QOR + $$\angle\ $$OQR = 180$$^{\circ\ }$$

$$=$$>  n$$^{\circ\ }$$ + n$$^{\circ\ }$$ + $$\angle\ $$OQR = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$OQR = 180$$^{\circ\ }$$- 2n$$^{\circ\ }$$

$$\angle\ $$OQP + $$\angle\ $$OQR = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$OQP + 180$$^{\circ\ }$$- 2n$$^{\circ\ }$$ = 180$$^{\circ\ }$$

$$=$$>  $$\angle\ $$OQP = 2n$$^{\circ\ }$$

In $$\triangle\ $$OPQ,

OQ = OP

$$=$$>  $$\angle\ $$OPQ = $$\angle\ $$OQP

$$=$$>  $$\angle\ $$OPQ = 2n$$^{\circ\ }$$

In $$\triangle\ $$OPR,

$$\angle\ $$POS is the external angle at point O

$$=$$>  $$\angle\ $$POS = $$\angle\ $$OPR + $$\angle\ $$ORP

$$=$$>  $$\angle\ $$POS = $$\angle\ $$OPQ + $$\angle\ $$ORP

$$=$$>  m$$^{\circ\ }$$ = 2n$$^{\circ\ }$$ + n$$^{\circ\ }$$

$$=$$>  m$$^{\circ\ }$$ = 3n$$^{\circ\ }$$

Hence, the correct answer is Option A