Given the problem, we need to find the value of the expression:
\(\frac{b - c}{r_1} + \frac{c - a}{r_2} + \frac{a - b}{r_3}\)
where \(r_1, r_2,\) and \(r_3\) are the exradii of \(\triangle ABC\) opposite to the vertices \(A, B,\) and \(C\), respectively.
To simplify this expression, the key concept involves the properties of exradii and the identity of triangle sides:
Recall the formula for the exradius \(r_1\) opposite to vertex \(A\):
\(r_1 = \frac{K}{s-a}\)
where \(K\) is the area of the triangle and \(s\) is the semi-perimeter.
Similarly, we have:
\(r_2 = \frac{K}{s-b}, \quad r_3 = \frac{K}{s-c}\)
Substitute these expressions into the given equation:
\(\frac{b-c}{\frac{K}{s-a}} + \frac{c-a}{\frac{K}{s-b}} + \frac{a-b}{\frac{K}{s-c}}\)
Simplifies to:
\(\frac{(b-c)(s-a)}{K} + \frac{(c-a)(s-b)}{K} + \frac{(a-b)(s-c)}{K}\)
Since each term has \(K\) in the denominator, multiply each term by \(K\) to eliminate it:
\((b-c)(s-a) + (c-a)(s-b) + (a-b)(s-c)\)
The above expands and simplifies using the identity for the semi-perimeter \((s = \frac{a+b+c}{2})\) and the cyclical properties of the expressions such that:
\(b(s-a) - c(s-a) + c(s-b) - a(s-b) + a(s-c) - b(s-c)\)
This simplifies to zero due to the cyclical and symmetric properties of triangle side lengths:
\(= b\left(\frac{a+b+c}{2} - a\right) - c\left(\frac{a+b+c}{2} - a\right) + c\left(\frac{a+b+c}{2} - b\right) - a\left(\frac{a+b+c}{2} - b\right) + a\left(\frac{a+b+c}{2} - c\right) - b\left(\frac{a+b+c}{2} - c\right)\)
Through simplification, this results in zero.
Thus, the value of the given expression is \(0\). Therefore, the correct option is:
0.