**Correct option (A) → (q); (B) → (p); (C) → (s); (D) → (r)**

(G_{3}A)_{2} = (G_{3}M) (G_{3}C)

Therefore,

So, (C) → (s).

Now in ΔAG_{1}C, we have

sin ∠CAM = asinC/2(AG_{1})

and in ΔBCG_{2}

_{sin }∠CBM = b sinC/2(BG_{2})

From Eq. (1), we have

a^{2} + b^{2} = 2c^{2} ......(3)

and from Eq. (2) and (3), we have

_{}

So, (B)→ (p)

Again, sin∠CAM + sin∠CBM is maximum when C = π/4

Also from Eq. (5), we have

c^{2 }= 2ab cos C

which implies

c^{2}/ab = 2 cos C = √2

So, (D)→ (r).