If \(\mathrm{A}(6,4,6), \mathrm{B}(12,4,0), \mathrm{C}(4,2,-2)\) are the vertices of triangle, then it's incentre is (A) \([(22 / 3),(10 / 3),(4 / 3)]\) (B) \([\\{(-22) / 3\\}(10 / 3),(4 / 3)]\) (C) \([(22 / 3),\\{(-10) / 3\\},(4 / 3)]\) (D) \([(22 / 3),(10 / 3),\\{(-4) / 3\\}]\)

First, use the distance formula to find the length of each side of the triangle: \(AB\), \(BC\), and \(AC\). Distance formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\) For sides AB, BC, and AC, we have: \(AB = \sqrt{(12 - 6)^2 + (4 - 4)^2 + (0 - 6)^2} = \sqrt{6^2 + 6^2} = 6\sqrt{2}\) \(BC = \sqrt{(4 - 12)^2 + (2 - 4)^2 + (-2 - 0)^2} = \sqrt{8^2 + 2^2 + 2^2} = 2\sqrt{17}\) \(AC = \sqrt{(4 - 6)^2 + (2 - 4)^2 + (-2 - 6)^2} = \sqrt{2^2 + 2^2 + 8^2} = 2\sqrt{17}\)

Now, apply the angle bisector theorem to get the ratio: The ratio in which angle bisector at vertex A divides side BC: \(\frac{AB}{AC} = \frac{6\sqrt{2}}{2\sqrt{17}} = \frac{3\sqrt{2}}{\sqrt{17}}\) Similarly for vertex B, the ratio in which angle bisector at vertex B divides side AC: \(\frac{AB}{BC} = \frac{6\sqrt{2}}{2\sqrt{17}} = \frac{3\sqrt{2}}{\sqrt{17}}\) Now, let's find the coordinates of the point on the line segment BC where the angle bisector at vertex A intercepts, and similarly for the angle bisector at vertex B on the line segment AC.

We can find the coordinates of these interception points on sides BC and AC using the section formula, given the ratio obtained above. Section formula: \((x, y, z) = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n}\right)\) For the angle bisector at vertex A, the point on side BC where it intercepts: \((x, y, z) = \left(\frac{\frac{3\sqrt{2}}{\sqrt{17}} \cdot 4 + \frac{\sqrt{17}}{\sqrt{17}} \cdot 12}{\frac{3\sqrt{2}}{\sqrt{17}} + \frac{\sqrt{17}}{\sqrt{17}}}, \frac{\frac{3\sqrt{2}}{\sqrt{17}} \cdot 2 + \frac{\sqrt{17}}{\sqrt{17}} \cdot 4}{\frac{3\sqrt{2}}{\sqrt{17}} + \frac{\sqrt{17}}{\sqrt{17}}}, \frac{\frac{3\sqrt{2}}{\sqrt{17}} \cdot (-2) + \frac{\sqrt{17}}{\sqrt{17}} \cdot 0}{\frac{3\sqrt{2}}{\sqrt{17}} + \frac{\sqrt{17}}{\sqrt{17}}}\right) = \left(\frac{22}{3}, \frac{10}{3}, \frac{4}{3}\right)\) Since the incentre is the point where the angle bisectors intersect, we can conclude that the incentre of the triangle is: \(\left(\frac{22}{3}, \frac{10}{3}, \frac{4}{3}\right)\) Therefore, the correct answer is (A).