In ∆ABC, if 3∠A = 4∠B = 6∠C then A : B : C = ?

LCM of 3, 4 and 6 = 12

3∠A = 4∠B = 6∠C      (Given)

Dividing throughout by 12, we get

$\frac{3 \angle \mathrm{A}}{12}=\frac{4 \angle \mathrm{B}}{12}=\frac{6 \angle \mathrm{C}}{12}$

$\Rightarrow \frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}$

Let $\frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}=k$, where $k$ is some constant

Then, ∠A = 4k,  ∠B = 3k, ∠C = 2k

∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2

Hence, the correct answer is option (b).