Given,
$$
\begin{aligned}
& \mathbf{a}=x^2 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
& \mathbf{b}=x \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathbf{a} \cdot \mathbf{b}>6 \\
& \Rightarrow\left(x^2 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\right)(x \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})>6 \\
& \Rightarrow x^2 \cdot x+(x)(-4)+3(2)>6
\end{aligned}
$$
$\left[\mathbf{a} \cdot \mathbf{b}=\mathbf{a}_1 \mathbf{a}_2+\mathbf{b}_1 \mathbf{b}_2+\mathbf{c}_1 \mathbf{c}_2\right.$, where
$$
\left.\mathbf{a}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}} \text { and } \mathbf{b}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+c_2 \hat{\mathbf{k}}\right]
$$
$$
\begin{array}{rrr}
\Rightarrow & x^3-4 x+6>6 \\
\Rightarrow & x^3-4 x+6-6>0 \\
\Rightarrow & x^3-4 x>0 \\
\Rightarrow & x\left(x^2-4\right)>0 \\
\Rightarrow & x(x-2)(x+2)>0
\end{array}
$$
Using wavy curve method critical points are
$$
x=0,2,-2
$$



$\therefore$ Solution is $x \in(-2,0) \cup(2, \infty)$.