Solution :

As the coefficient matrix is not diagonally dominant we rearrange the equations

$$ \begin{array}{l} 28 x+4 y-z=32 \\ 2 x+17 y+4 z=35 \\ x+3 y+10 z=24 \end{array} $$

Since the diagonal elements are dominant in the coefficient matrix, we write $\mathrm{x},\mathrm{y}, \mathrm{z}$ as follows:

$$ \begin{aligned} x &=\frac{1}{28} 32-4 y+z \\ y &=\frac{1}{7} 35-2 x-4 z \\ z &=\frac{1}{10} 24-x-3 y \end{aligned} $$ Let the initial values be $y_{0}=0, z_{0}=0$

First Iteration : $$ \begin{aligned} x_{1} &=\frac{1}{28} 32-4 y_{0}+z_{0}=1.1429 ; \\ y_{1} &=\frac{1}{7} 35-2 x_{1}-4 z_{0}=1.9244 ; \\ z_{1} &=\frac{1}{10} 24-x_{1}-3 y_{1}=1.7084 \end{aligned} $$

Second Iteration : $$ \begin{aligned} x_{2} &=\frac{1}{28} 32-4 y_{1}+z_{1}=0.929 ; \\ y_{2} &=\frac{1}{7} 35-2 x_{2}-4 z_{1}=0.929 ;\\ z_{2} &=\frac{1}{10} 24-x_{2}-3 y_{2}=1.8428 \end{aligned} $$ Third Iteration : $$ \begin{aligned} x_{3} &=\frac{1}{28} 32-4 y_{2}+z_{2}=0.9876 ; \\ y_{3} &=\frac{1}{7} 35-2 x_{3}-4 z_{2}=1.509 ; \\ z_{3} &=\frac{1}{10} 24-x_{3}-3 y_{3}=1.8485 \end{aligned} $$ Fourth Iteration : $$ \begin{aligned} x_{4} &=\frac{1}{28} 32-4 y_{3}+z_{3}=0.9933 ; \\ y_{4} &=\frac{1}{7} 35-2 x_{4}-4 z_{3}=1.507 ; \\ z_{4} &=\frac{1}{10} 24-x_{4}-3 y_{4}=1.8486 \end{aligned} $$ Fifth Iteration : $$ \begin{aligned} x_{5} &=\frac{1}{28} 32-4 y_{4}+z_{4}=0.9936 ; \\ y_{5} &=\frac{1}{7} 35-2 x_{5}-4 z_{4}=1.507 ; \\ z_{5} &=\frac{1}{10} 24-x_{5}-3 y_{5}=1.8485 \end{aligned} $$

Here Fourth Iteration $=$ Fifth Iteration Hence $x=0.993 ; \ y=1.507 ; \ z=1.848$

correct to three decimal places.