Let $\frac{1}{x}=p, \frac{1}{y} q, \frac{1}{z}=r$

Then the given system of equations is as follows:

$2 p+3 q+10 r=4$

$4 p-6 q+5 r=1$

$6 p+9 q+20 r=2$

This system can be written in the form of $A X=B$, where

$A=\left[\begin{array}{ccc}2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20\end{array}\right], X=\left[\begin{array}{l}p \\ q \\ r\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right]$

Now, $|A|=2(120-45)-3(-80-30)+10(36+36)$

$=150+330+720$

$=1200$

Thus, $A$ is non-singular. Therefore, its inverse exists. Now, $A_{11}=75, A_{12}=110, A_{13}=72$

$A_{21}=150, A_{22}=100, A_{23}=0$

$A_{31}=75, A_{32}=30, A_{33}=-24$

$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$

$=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]$

Now,

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}p \\ q \\ r\end{array}\right]=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right]$

$=\frac{1}{1200}\left[\begin{array}{c}300+150+150 \\ 440-100+60 \\ 288+0-48\end{array}\right]$

$=\frac{1}{1200}\left[\begin{array}{l}600 \\ 400 \\ 240\end{array}\right]=\left[\begin{array}{l}\frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5}\end{array}\right]$

$\therefore p=\frac{1}{2}, q=\frac{1}{3},$ and $r=\frac{1}{5}$

Hence, $x=2, y=3$ and $z=5$