If one root of the equation $x^2 - kx + k = 0$ exceeds the other by $2\sqrt{3}$, then which one of the following is a value of $k$?

Let the roots of the equation $x^2 - kx + k = 0$ be $\alpha$ and $\beta$.

Given that one root exceeds the other by $2\sqrt{3}$, we have $|\alpha - \beta| = 2\sqrt{3}$.

Squaring both sides, we get $(\alpha - \beta)^2 = 12$.

Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$, we can substitute the sum and product of the roots.

From the given quadratic equation, $\alpha + \beta = k$ and $\alpha\beta = k$.

Substituting these values, we get $k^2 - 4k = 12$.

$k^2 - 4k - 12 = 0$

$(k - 6)(k + 2) = 0$

Thus, $k = 6$ or $k = -2$.

Among the given options, $6$ is a value of $k$.

Answer: $6$