In a triangle ABC with usual notations, if $3 a=\mathrm{b}+\mathrm{c}$, then $\cot \frac{B}{2} \cdot \cot \frac{C}{2}=$

The product 【cot(B/2) × cot(C/2)】 is evaluated using the half-angle formulas in a triangle:
$\cot\frac{B}{2} = \sqrt{\frac{s(s − b)}{(s − a)(s − c)}}$ and $\cot\frac{C}{2} = \sqrt{\frac{s(s − c)}{(s − a)(s − b)}}$, where $s = \frac{a + b + c}{2}$ is the semi-perimeter.

Multiplying the expressions gives:
$\cot\frac{B}{2} \cdot \cot\frac{C}{2} = \sqrt{\frac{s(s − b)}{(s − a)(s − c)}} \cdot \sqrt{\frac{s(s − c)}{(s − a)(s − b)}} = \sqrt{\frac{s^2}{(s − a)^2}} = \frac{s}{s − a}.$

Given b + c = 3a, the semi-perimeter simplifies to $s = \frac{a + b + c}{2} = \frac{a + 3a}{2} = 2a$.
Substitute into the product:
$\cot\frac{B}{2} \cdot \cot\frac{C}{2} = \frac{2a}{2a − a} = \frac{2a}{a} = 2.$

Thus, the value is $\boxed{C}$ from the options provided.