If the line $$x - 1 = 0$$ is a directrix of the hyperbola $$kx^2 - y^2 = 6$$, then the hyperbola passes through the point

We need to find which point the hyperbola passes through, given that $$x = 1$$ is a directrix of $$kx^2 - y^2 = 6$$.

$$kx^2 - y^2 = 6 \implies \frac{x^2}{6/k} - \frac{y^2}{6} = 1$$

This is of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with $$a^2 = \frac{6}{k}$$ and $$b^2 = 6$$.

$$b^2 = a^2(e^2 - 1)$$

$$6 = \frac{6}{k}(e^2 - 1)$$

$$k = e^2 - 1$$

$$e^2 = k + 1$$

For a hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$, the directrices are at $$x = \pm \frac{a}{e}$$.

Given that $$x = 1$$ is a directrix:

$$\frac{a}{e} = 1 \implies a = e$$

$$a^2 = e^2$$

$$\frac{6}{k} = k + 1$$

$$6 = k(k + 1) = k^2 + k$$

$$k^2 + k - 6 = 0$$

$$(k + 3)(k - 2) = 0$$

$$k = 2$$ (taking the positive value since $$k > 0$$ for a valid hyperbola)

$$2x^2 - y^2 = 6$$

Option A: $$(-2\sqrt{5}, 6)$$: $$2(20) - 36 = 40 - 36 = 4 \neq 6$$

Option B: $$(-\sqrt{5}, 3)$$: $$2(5) - 9 = 10 - 9 = 1 \neq 6$$

Option C: $$(\sqrt{5}, -2)$$: $$2(5) - 4 = 10 - 4 = 6 = 6$$ $$\checkmark$$

Option D: $$(2\sqrt{5}, 3\sqrt{6})$$: $$2(20) - 54 = 40 - 54 = -14 \neq 6$$

Therefore, the correct answer is Option C: $$(\sqrt{5}, -2)$$.