If (a + b) : (b + c) : (c + a) = 3 : 6 : 7 and 3a + 5b = 121 , then b…

a + b = 3R, b + c = 6R, c + a =7R

Adding all of above, we get:

2(a + b + c) = 16R

(a + b + c) = 8R      ...............(i)

b + c = 6R   ...........(ii)

(i) - (ii)

a = 2R, similarly b = 1R,  c = 5R

Now,

⇒  3a +  5b = 66

⇒  (3 × 2R) +  (5 × 1R) = 121

⇒  6R  + 5R = 121

⇒  11R = 121

⇒  1R = 11

 Here,

b  = 1R = 1 × 11 = 11