We know that the formula to calculate the area of triangle with vertices $$\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right) \&\left(\mathrm{x}_3, \mathrm{y}_3\right)$$ is

$$\frac{1}{2}\left|\mathrm{x}_1\left(\mathrm{y}_2-\mathrm{y}_3\right)+\mathrm{x}_2\left(\mathrm{y}_3-\mathrm{y}_1\right)+\mathrm{x}_3\left(\mathrm{y}_1-\mathrm{y}_2\right)\right|$$

Using the above formula, the area of the triangle, if any, formed by the points $$A(4,6), B(8,10)$$ $$\& C(12,14)$ is $\frac{1}{2}|4(10-14)+8(14-6)+12(6-10)|=\frac{1}{2}|-16+64-48|=\frac{1}{2} \times 0=0$$

The area being zero means the three points $$\mathrm{A}, \mathrm{B} \& \mathrm{C}$$ are collinear i.e. in a straight line.

Since the points are on a straight line, the slope found using any two of the three points should be the same.

Hence all three options (A), (B) \& (C) are correct.