Step-by-step Solution:

Statement A: "If \( a \cos A = b \cos B \), then the triangle is isosceles or right-angled." - Analysis: Using the Law of Cosines, we know: \[ a \cos B + b \cos A = c \] If \( a \cos A = b \cos B \), then the triangle must satisfy this condition. This can happen if: 1. The triangle is isosceles (\( a = b \)), or 2. The triangle is right-angled (with \( C = 90^\circ \)). Thus, Statement A is true. --- \[Statement B:\] "If in a triangle \( ABC \), \( \cos A \cos B + \sin A \sin B \sin C = 1 \), then the triangle is isosceles right-angled." - Analysis: The given equation can be rewritten using trigonometric identities: \[ \cos A \cos B + \sin A \sin B \sin C = \cos(A - B) + \sin A \sin B \sin C = 1 \] For this equation to hold, the only possibility is: \[ \cos(A - B) = 1 \quad \text{and} \quad \sin A \sin B \sin C = 0 \] This implies: 1. \( A = B \) (isosceles), and 2. \( \sin C = 1 \) (right-angled, \( C = 90^\circ \)). Thus, Statement B is true. \[Statement C:\] "If the ex-radii \( r_1, r_2, r_3 \) of \( \triangle ABC \) are in harmonic progression (HP), then its sides are not in arithmetic progression (AP)." - Analysis: The ex-radii \( r_1, r_2, r_3 \) are given by: \[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \] If \( r_1, r_2, r_3 \) are in HP, then \( \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \) are in AP. This implies: \[ \frac{1}{r_1} + \frac{1}{r_3} = \frac{2}{r_2} \] Substituting the expressions for \( r_1, r_2, r_3 \): \[ \frac{s - a}{\Delta} + \frac{s - c}{\Delta} = \frac{2(s - b)}{\Delta} \] Simplifying: \[ (s - a) + (s - c) = 2(s - b) \] \[ 2s - a - c = 2s - 2b \] \[ -a - c = -2b \] \[ a + c = 2b \] This means the sides \( a, b, c \) are in AP. Thus, Statement C is false. --- Final Answer: - Statement A: True - Statement B: True - Statement C: False The correct option is: \[ \boxed{A} \]