In a triangle $A B C$, with usual notations. $\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}$ Then $\angle \mathrm{A}=$

Given the equation:

$\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{b c} + \frac{b}{c a}$

Apply the cosine rule to express trigonometric terms:

$\cos A = \frac{b^2 + c^2 - a^2}{2 b c}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2 a c}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2 a b}$

Substitute into the left-hand side:

\begin{align*} \text{LHS} &= \frac{2}{a} \cdot \frac{b^2 + c^2 - a^2}{2 b c} + \frac{1}{b} \cdot \frac{a^2 + c^2 - b^2}{2 a c} + \frac{2}{c} \cdot \frac{a^2 + b^2 - c^2}{2 a b} \\ &= \frac{b^2 + c^2 - a^2}{a b c} + \frac{a^2 + c^2 - b^2}{2 a b c} + \frac{a^2 + b^2 - c^2}{a b c} \end{align*}

Combine terms over a common denominator of $2 a b c$:

$\frac{2(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2)}{2 a b c}$

Simplify the numerator:

$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = a^2 + 3b^2 + c^2$

Thus, $\text{LHS} = \frac{a^2 + 3b^2 + c^2}{2 a b c}$

The right-hand side simplifies to:

$\frac{a}{b c} + \frac{b}{c a} = \frac{a^2 + b^2}{a b c}$

Equating both sides:

$\frac{a^2 + 3b^2 + c^2}{2 a b c} = \frac{a^2 + b^2}{a b c}$

Multiply through by $2 a b c$:

$a^2 + 3b^2 + c^2 = 2(a^2 + b^2)$

Simplify:

$a^2 + 3b^2 + c^2 = 2a^2 + 2b^2 \Rightarrow b^2 + c^2 = a^2$

This is a Pythagorean relation, indicating that angle $A$ is a right angle.

Conclusion: $\angle A = 90^\circ = \frac{\pi}{2}$

The correct choice is $\boxed{\text{C}}$.