• # question_answer The interval of a for which (a, a2, a) and (3, 2,1) lies on same side of$x+y-4z+2=0$is A)  $(\infty ,1]\cup (2,\infty ]$       B)  $[-\infty ,1)\cup [2,\infty )$ C)  $(-\infty ,2]\cup (3,\infty ]$     D)  $(-\infty ,2)\cup [3,\infty )$

Here, it is given that (a, a2, a) and (3, 2, 1) lies on same side of x. + y - 4z + 2 = 0 $\therefore$ (a + a2-4a + 2) (3 + 2 - 4 + 2) > 0 $\Rightarrow$                               $({{a}^{2}}-3a+2)>0$                                 $(a-1)(a-2)>0$                                 $a\in (-\infty ,1]\cup (2,\infty ]$