If $\vec{a}=x \hat{i}+(x-1) \hat{j}+\hat{k}$ and $\vec{b}=(x+1) \hat{i}+\hat{j}+a \hat{k}$ always make an acute angle with each other for every value of $x \in R$, then:

$a \in(2, \infty)$

$\vec{a} . \vec{b}=(x \hat{i}+(x-1) \hat{j}+\hat{k}) . ((x+1) \hat{i}+\hat{j}+a \hat{k})$

$=x(x+1)+x-1+a$

$=x^2+2 x+a-1$

We must have

$\vec{a} . \vec{b}>0 ~\forall~ x \in R$

$\Rightarrow x^2+2 x+a-1>0 ~\forall~ x \in R$

⇒ 4 – 4(a – 1) < 0

⇒ a > 2

Hence (2) is correct answer.