120. Since $\triangle A S R \sim \triangle A B C$, we have $\frac{x}{a}=\frac{h_{a}-x}{h_{a}}$, i.e., $x=\frac{a h_{a}}{a+h_{a}}$. Similarly, $y=\frac{b h_{b}}{b+h_{b}}, z=\frac{c h_{c}}{c+h_{c}}$. Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{a}{2 S}+\frac{b}{2 S}+\frac{c}{2 S}$. Thus, the original inequality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{(2+\sqrt{3}) p}{2 S}$ is equivalent to proving
$$\begin{array}{l} a+b+c+\frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c} \leqslant \frac{(2+\sqrt{3})(a+b+c)}{2} \Leftrightarrow \\ \frac{2 S}{a}+\frac{2 S}{b}+\frac{2 S}{c} \leqslant \frac{\sqrt{3}(a+b+c)}{2} \end{array}$$

By Heron's formula, it is equivalent to proving
$$\begin{array}{l} \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leqslant \\ \sqrt{3}(a+b+c) \Leftrightarrow \\ \sqrt{(-a+b+c)(a-b+c)(a+b-c)}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leqslant \sqrt{3(a+b+c)} \end{array}$$

Let $u=-a+b+c, v=a-b+c, w=a+b-c$, then the inequality becomes
$$\begin{array}{l} 2 u v w\left(\left(\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u}\right) \leqslant \sqrt{3(u+v+w)} \Leftrightarrow\right. \\ 4\left(\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u}\right)^{2} \leqslant 3\left(\frac{1}{u v}+\frac{1}{v w}+\frac{1}{w u}\right) \end{array}$$

By the AM-GM inequality and Cauchy-Schwarz inequality, we get
$$\begin{array}{l} 4\left(\frac{1}{u+v}+\frac{1}{v+w}+\frac{1}{w+u}\right)^{2}=\left(\frac{2}{u+v}+\frac{2}{v+w}+\frac{2}{w+u}\right)^{2} \leqslant \\ \left(\frac{1}{\sqrt{u v}}+\frac{1}{\sqrt{v w}}+\frac{1}{\sqrt{w u}}\right)^{2} \leqslant \\ (1+1+1)\left(\frac{1}{u v}+\frac{1}{v w}+\frac{1}{w u}\right)= \\ 3\left(\frac{1}{u v}+\frac{1}{v w}+\frac{1}{w u}\right) \end{array}$$