Step 1: Substitute variables to linearise the equations. Let $a = \frac{1}{x}$, $b = \frac{1}{y}$, and $c = \frac{1}{z}$. The system becomes:

$$2a + 3b + 10c = 4$$ $$4a - 6b + 5c = 1$$ $$6a + 9b - 20c = 2$$

Step 2: Write the system in matrix form $AX = B$.

$$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}, B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$$

Step 3: Calculate the determinant $|A|$. Expanding along Row 1:

$$|A| = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)$$ $$|A| = 2(75) - 3(-110) + 10(72)$$ $$|A| = 150 + 330 + 720 = 1200$$

Since $|A| \neq 0$, the system has a unique solution.

Step 4: Find the cofactors and adjoint of A. Cofactors are calculated as follows:

$$C_{11} = 75, C_{12} = 110, C_{13} = 72$$ $$C_{21} = 150, C_{22} = -100, C_{23} = 0$$ $$C_{31} = 75, C_{32} = 30, C_{33} = -24$$ $$adj(A) = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$$

Step 5: Calculate $X = A^{-1}B$.

$$A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$$ $$X = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$$ $$X = \frac{1}{1200} \begin{bmatrix} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}$$

Step 6: Solve for x, y, and z. Since $a = 1/2, b = 1/3, c = 1/5$, we have:

$$x = \frac{1}{a} = 2, y = \frac{1}{b} = 3, z = \frac{1}{c} = 5$$