The vectors  $(x^2-1)\overline{i}+2(x^2-1)\overline{j}-3(x^2-1)\overline{k},(2x^2-1)\overline{i}+(2x^2+1)\overline{j}+x^2\overline{k},$ and  $(3x^2+2)\overline{i}+(x^2+4)\overline{j}+(x^2+1)\overline{k}$  are non-coplanar. The number of real values that x cannot take is

The scale triple product of the three given vectors is $\left|\begin{array}{ccc}{x}^2-1& 2(x^2-1)& -3(x^2-1)\\ 2x^2-1& 2x^2+1& x^2\\ 3x^2+2& x^2+4& x^2+1\end{array}\right|$

$=(x^2-1)\left|\begin{array}{ccc}1& 2& -3\\ 2x^2-1& 2x^2+1& x^2\\ 3x^2+2& x^2+4& x^2+1\end{array}\right|=(x^2-1)\left|\begin{array}{ccc}1& 2& 0\\ 2x^2-1& 2x^2+1& x^2\\ 3x^2+2& x^2+4& 5x^2+7\end{array}\right|$

$(u\sin g C_3\to C_3+C_2+C_1)$

$=(x^2-1)\left|\begin{array}{ccc}1& 0& 0\\ 2x^2-1& -2x^2+3& 5x^2\\ 3x^2+2& -5x^2& 5x^2+7\end{array}\right| (u\sin g C_2\to C_2=2C_1)$

$=(x^2-1)(15x^4+x^2+21)$

Vector are non-coplanar  $\Rightarrow$ Scalar triple product 

$\ne 0\Rightarrow x^2-1\ne 0(\because 15x^4+x^2+21\ne 0)\Rightarrow x\ne \pm 1$