We first write the equation as

$x_1 = \dfrac {1}{10} (12 - x_2 - x_3) \ \cdots \ \cdots (1)$

$x_2 = \dfrac {1}{10} (13 - 2x_1 - x_3) \ \cdots \ \cdots (2)$

$x_3 = \dfrac {1}{10} (14 - 2x_1 - 2x_2) \ \cdots \ \cdots (3)$

(i) First Iteration:

We start with the approximation $x_2 = 0, \ x_3 =0$ and then we get from (1)

$\therefore x_{1,1} = \dfrac {12}{10} = 1.2$

We use this approximation to find $x_2$ i.e. we put $x_1 = 1.2, \ x_3 = 0$ in (2)

$\therefore x_{2,1} = \dfrac {1}{10} \big [ 13-2 (1.2)\big ] = 1.06$

We use these values of $x_1$ and $x_2$ to find $x_3$

$\therefore x_{3,1} = \dfrac {1}{10} \big [ 14-2 (1.2) - 2 (1.06) \big ] = 0.948$

(ii) Second Iteration:

We use the latest of $x_2$ and $x_3$ to find $x_1$ i,e,

We put $x_2 = 1.06, \ x_3 = 0.948$ and then we get from (1)

$\therefore x_{1,2} = \dfrac {1}{10} \big [ 12-1.06 - 0.948 \big ] = 0.9992$

We use this approximation to find $x_2$ i.e. we put

$x_1 = 0.9992, \ x_3 = 0.948$ in (2)

$\therefore x_{2,2} = \dfrac {1}{10} \big [ 13-2 (0.9992) - 0.948 \big ] = 1.00536$

We use these values of $x_1$ and $x_2$ to find $x_3$

$\therefore x_{3,2}= \dfrac {1}{10} \big [ 14 -2 (0.9992) - 2 (1.00536) \big ] = 0.999088$

(iii) Third Iteration:

We use the latest values of $x_2$ and $x_3$ to find $x_1$ i.e.

We put $x_2 = 1.00536, \ x_3 = 0.999088$ and then we get from (1)

$\therefore x_{1,3} = \dfrac {1}{10} \big [ 12-1.00536 - 0.999088 \big ] = 0.9995552$

We use this approximation to find $x_2$ i.e. we put

$x_1 = 0.9995552,$

$x_3 = 0.999088$ in (2)

$\therefore x_{2,3} = \dfrac {1}{10} \big [ 13-2 (0.9995552) - 0.999088 \big ] = 1.00018$

We use these of $x_1$ and $x_2$ to find $x_3$

$\therefore x_{3,3} = \dfrac {1}{10} \big [ 14 - 2 (0.9995552) - 2 (1.00018) \big ] = 1.00052$

Since, the second and third iteration give the same values.

$x_1 = 1, x_2 = 1, x_3=1$