Find dual from primal conversion
MIN z = x1  3x2  2x3
subject to
3x1  x2 + 2x3 <= 7
2x1  4x2 >= 12
4x1 + 3x2 + 8x3 = 10
and x1,x2 >= 0 and x3 unrestricted in sign
Solution:
Primal is (Solution stpes of Primal by BigM method)
MIN `z_x`  `=`  ``  ``  `x_1`  `  `  `3`  `x_2`  `  `  `2`  `x_3` 

subject to 
``  `3`  `x_1`  `  `  ``  `x_2`  ` + `  `2`  `x_3`  ≤  `7`  ``  `2`  `x_1`  `  `  `4`  `x_2`     ≥  `12`  `  `  `4`  `x_1`  ` + `  `3`  `x_2`  ` + `  `8`  `x_3`  =  `10` 

and `x_1,x_2 >= 0; ``x_3` unrestricted in sign 
all `<=` constraints can be converted to `>=` type by multipling both sides by 1
MIN `z_x`  `=`  ``  ``  `x_1`  `  `  `3`  `x_2`  `  `  `2`  `x_3` 

subject to 
`  `  `3`  `x_1`  ` + `  ``  `x_2`  `  `  `2`  `x_3`  ≥  `7`  ``  `2`  `x_1`  `  `  `4`  `x_2`     ≥  `12`  `  `  `4`  `x_1`  ` + `  `3`  `x_2`  ` + `  `8`  `x_3`  =  `10` 

and `x_1,x_2 >= 0; ``x_3` unrestricted in sign 
The `x_3` variable in the primal is unrestricted in sign, therefore the `3^(rd)` constraint in the dual shall be equality.
Since `3^(rd)` constraint in the primal is equality, the corresponding dual variable `y_3` will be unrestricted in sign.
Dual is (Solution stpes of Dual by BigM method)
MAX `z_y`  `=`  `  `  `7`  `y_1`  ` + `  `12`  `y_2`  ` + `  `10`  `y_3` 

subject to 
`  `  `3`  `y_1`  ` + `  `2`  `y_2`  `  `  `4`  `y_3`  ≤  `1`  ``  ``  `y_1`  `  `  `4`  `y_2`  ` + `  `3`  `y_3`  ≤  `3`  `  `  `2`  `y_1`     ` + `  `8`  `y_3`  =  `2` 

and `y_1,y_2 >= 0; ``y_3` unrestricted in sign 
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then