The foot of perpendicular of the point $$(2, 0, 5)$$ on the line $$\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$$ is $$(\alpha, \beta, \gamma)$$. Then, which of the following is NOT correct?

$$\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda$$

This gives the coordinates of any general point $$F(\alpha, \beta, \gamma)$$ on the line:

$$\alpha = 2\lambda - 1$$

$$\beta = 5\lambda + 1$$

$$\gamma = -\lambda - 1$$

The given external point is $$P(2, 0, 5)$$. The direction ratios of the line segment connecting $$P$$ to the general foot $$F$$ are: $$\langle \alpha - 2, \ \beta - 0, \ \gamma - 5 \rangle$$
$$a_2 = (2\lambda - 1) - 2 = 2\lambda - 3$$

$$b_2 = (5\lambda + 1) - 0 = 5\lambda + 1$$

$$c_2 = (-\lambda - 1) - 5 = -\lambda - 6$$

The direction ratios of the given line are $$\langle a_1, b_1, c_1 \rangle = \langle 2, 5, -1 \rangle$$. Since $$\vec{PF}$$ is perpendicular to the line:

$$2(2\lambda - 3) + 5(5\lambda + 1) - 1(-\lambda - 6) = 0$$

$$30\lambda + 5 = 0 \implies \lambda = -\frac{5}{30} = -\frac{1}{6}$$

$$\alpha = 2\left(-\frac{1}{6}\right) - 1 = -\frac{1}{3} - 1 = -\frac{4}{3}$$

$$\beta = 5\left(-\frac{1}{6}\right) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$$

$$\gamma = -\left(-\frac{1}{6}\right) - 1 = \frac{1}{6} - 1 = -\frac{5}{6}$$

$$\frac{\beta}{\gamma} = \frac{1/6}{-5/6} = -\frac{1}{5} = -0.2$$

Hence, option C is incorrect.