If \(\mathrm{A}(1,2)\) and \(\mathrm{B}(6,2), 3 \mathrm{AB}=2 \mathrm{BC}\) and \(\mathrm{A}-\mathrm{B}-\mathrm{C}\) the value of \(\mathrm{C}\) can be (a) \([-(3 / 2),(3 / 3)]\) (b) \([(27 / 2), 2]\) (c) \([-(27 / 2), 2]\) (d) \([(27 / 2),-2]\)

To find AB, we can use the distance formula. The distance between A(x1, y1) and B(x2, y2) is given by: $$AB = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$$ Using the coordinates of A(1, 2) and B(6, 2), we get: $$AB = \sqrt{(6 - 1)^2 + (2 - 2)^2}$$ $$AB = \sqrt{5^2}$$ $$AB = 5$$

We know from the problem that 3AB = 2BC. Substitute AB with the value that we found to get the distance BC: $$3 \times 5 = 2 \times BC$$ $$15 = 2 \times BC$$ Divide both sides by 2 to find the value of BC: $$BC = \frac{15}{2}$$

Since A, B, and C lie on a straight line, and we know BC, we can find the movement from point B to point C to get the coordinates of point C. Since only the x-coordinates of points A and B differ, we can assume that the x-coordinate of point C will also differ while the y-coordinate will remain constant. The distance between the x-coordinates will follow the ratio given. Calculate the movement in x-coordinate (Δx): $$Δx = \frac{BC}{AB}(x_B - x_A)$$ $$Δx = \frac{15}{2 \times 5}(6 - 1)$$ $$Δx = \frac{15}{10}(5)$$ $$Δx = \frac{3}{2}(5)$$ $$Δx = \frac{15}{2}$$ Now, calculate the coordinates of point C (C_x, C_y): $$C_x = x_B + Δx$$ $$C_x = 6 + \frac{15}{2}$$ $$C_x = \frac{27}{2}$$ Since the y-coordinates of points A and B are the same, the y-coordinate of point C remains the same: $$C_y = 2$$ Thus, the coordinates of point C are \(\left(\frac{27}{2}, 2\right)\). The correct answer is (b).