### Solution:

Given that 10 years ago father was 12 times as old as his son and after 10 years he will be two times as old as his son. We need to find the present age of father and son.Let x be the present age of the son and y be the present age of the father.

10 years ago father was 12 times old as his son, therefore, we have,

$\n \n \n \n \n \n \n \n y\u221210=12(x\u221210)\n \n \n \n \n \u21d2y\u221210=12x\u2212120\n \n \n \n \n \u21d212x\u2212y=120\u221210\n \n \n \n \n \n \n \n \n \n \u21d212x\u2212y=110\u2026\u2026\u2026\u2026\u2026\u2026.(1)\n \n \n \n \n $

After 10 years father will be twice as old as his son, thus, we have,

$\n \n \n \n y+10=2(x+10)\n \n \n \n \n \u21d2y+10=2x+20\n \n \n \n \n \u21d22x\u2212y=\u221220+10\n \n \n \n \n \u21d22x\u2212y=\u221210\u2026\u2026\u2026\u2026.(2)\n \n \n \n \n $

Subtract equations (2) from (1), we have,

$\n \n \n \n \n \n \n \n 12x\u2212y\u2212(2x\u2212y)\n \n \n \n \n 12x\u2212y\u22122x+y\n \n \n \n \n \n \n \n \n \n 110\u2212(\u221210)\n \n \n \n \n =110+10\n \n \n \n \n \n \n \n \n \n \u21d210x\n \n \n \n =120\n \n \n \n \n \n \u21d2x\n \n \n \n =12\n \n \n \n \n $

Substitute the value of x in equation (2), we have,

$\n \n \n \n \n \n 2(12)\u2212y\n \n \n \n =\u221210\n \n \n \n \n \n \u21d224\u2212y\n \n \n \n =\u221210\n \n \n \n \n \n \u21d2y\n \n \n \n =34\n \n \n \n \n $

Thus, the present age of son is 12 years and the present age of father is 34 years.

Therefore, option 2 is correct.