Solution:
Answer: $\sqrt{19}$

From the identity $\tan \frac{u}{2} = \frac{\sin u}{1 + \cos u}$, the conditions work out to $3 \tan \frac{w}{2} = 4 \tan \frac{x}{2} = 5 \tan \frac{y}{2} = 6 \tan \frac{z}{2} = k$. Let $a = \tan \frac{w}{2}$, $b = \tan \frac{x}{2}$, $c = \tan \frac{y}{2}$, and $d = \tan \frac{z}{2}$. Using the identity $\tan (M + N) = \frac{\tan M + \tan N}{1 - \tan M \tan N}$, we obtain
$$\begin{aligned} \tan \left(\frac{w + x}{2} + \frac{y + z}{2}\right) & = \frac{\tan \left(\frac{w + x}{2}\right) + \tan \left(\frac{y + z}{2}\right)}{1 - \tan \left(\frac{w + x}{2}\right) \tan \left(\frac{y + z}{2}\right)} \\ & = \frac{\frac{a + b}{1 - a b} + \frac{c + d}{1 - c d}}{1 - \left(\frac{a + b}{1 - a b}\right)\left(\frac{c + d}{1 - c d}\right)} \\ & = \frac{a + b + c + d - a b c - a b d - b c d - a c d}{1 + a b c d - a b - a c - a d - b c - b d - c d} \end{aligned}$$
Because $x + y + z + w = 2\pi$, we get that $\tan \left(\frac{x + y + z + w}{2}\right) = 0$ and thus $a + b + c + d = a b c + a b d + b c d + a c d$. Substituting $a, b, c, d$ corresponding to the variable $k$, we obtain that $k^{3} - 19 k = 0$. Therefore, $k$ can be only $0$, $\sqrt{19}$, $-\sqrt{19}$. However, $k = 0$ is impossible as $w, x, y, z$ will all be $0$. Also, $k = -\sqrt{19}$ is impossible as $w, x, y, z$ will exceed $\pi$. Therefore, $k = \sqrt{19}$.