If $$a=2b=3c$$​ and $$a+b+c=121$$​ then $$\sqrt{a^2+b^2+c^2}$$​ is:

Correct option is B

Given:

a = 2b = 3c, and a + b + c = 121

Solution:

a +$$\frac{a}{2} + \frac{a}{3} = \frac{6a + 3a + 2a}{6} = \frac{11a}{6}$$​

​$$\frac{11a}{6} = 121$$​

​$$a = \frac{121 \times 6}{11} = 66$$​

Now,

b =$$\frac{66}{2} = 33,$$​​

c =$$\frac{66}{3} = 22$$​​

​$$\sqrt{a^2 + b^2 + c^2} = \sqrt{66^2 + 33^2 + 22^2} = \sqrt{4356 + 1089 + 484} = \sqrt{5929} = 77$$​