The equation for the acute angle bisector of two planes is established. The normal vectors of the planes \(P_1: x - 2y - 2z + 1 = 0\) and \(P_2: 2x - 3y - 6z + 1 = 0\) are \( \mathbf{n_1} = (1, -2, -2) \) and \( \mathbf{n_2} = (2, -3, -6) \) respectively. The magnitudes of these normal vectors are \( |\mathbf{n_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = 3 \) and \( |\mathbf{n_2}| = \sqrt{2^2 + (-3)^2 + (-6)^2} = 7 \). The general form for the plane bisectors is \( \frac{x - 2y - 2z + 1}{3} = \pm \frac{2x - 3y - 6z + 1}{7} \). To determine the acute angle bisector, the dot product of the normal vectors is calculated: \( \mathbf{n_1} \cdot \mathbf{n_2} = (1)(2) + (-2)(-3) + (-2)(-6) = 2 + 6 + 12 = 20 \). Since \( \mathbf{n_1} \cdot \mathbf{n_2}>0 \), the negative sign is used for the acute bisector equation: \( \frac{x - 2y - 2z + 1}{3} = -\frac{2x - 3y - 6z + 1}{7} \). Multiplying both sides by 21 yields \( 7(x - 2y - 2z + 1) = -3(2x - 3y - 6z + 1) \). Expanding and rearranging terms results in the equation of the acute angle bisector: \( 13x - 23y - 32z + 10 = 0 \). Verification by substituting the coordinates of the point \( (-2, 0, -\frac{1}{2}) \) into this equation confirms that the point lies on the acute angle bisector.