The interval of a for which $(a,a^2,a)$ and $(3, 2, 1)$ lies on same side of $x+y−4z+2=0$ is:

Question

I found this question in a test and the answer said that $\infty$ should be included in the range. Is this possible?

Edit: The question is that, should I include $\infty$ in the range i.e. [-$\infty$, 1) U ( 2, $\infty$] or (-$\infty$, 1) U ( 2, $\infty$)

Answer 1

The plane divides space into three nonoverlapping regions. Points on the plane make the expression $x+y-4z+2$ to vanish. Points on one side of the plane make it positive, and points on the other side make it negative.

Thus for your points to be on the same side, they must have the same sign when their coordinates are substituted into the linear expression above. Putting the constant coordinates, we see that we get $3+2-4+2>0.$ It follows that we must have $a+a^2-4a+2>0$ as well. This is the inequality you need to solve for $a.$

Answer 2

Just solve $$(a+a^2-4a+2)(3+2-4\cdot1+2)>0.$$