The values of  $\mathrm{c}$ so that for all real $\mathrm{x}$, the vectors

 $\mathrm{cx}\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+3\widehat{\mathrm{k}} \mathrm{and} \mathrm{x}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\mathrm{cx}\widehat{\mathrm{k}}$ , 

make an obtuse angle with each other. Then which of the following is true?

Given $\mathrm{a}=\mathrm{cx}\widehat{\mathrm{i}}-6\widehat{\mathrm{j}}+3\widehat{\mathrm{k}} \mathrm{and} \mathrm{b}=\mathrm{x}\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}+2\mathrm{cx}\widehat{\mathrm{k}}$
$\overline{\mathrm{a}}$ and  $\overline{\mathrm{b}}$ makes an obtuse angle with each other
$\begin{array}{ll}\therefore & \cos \mathrm{\theta }=\frac{\mathrm{a}\cdot \mathrm{b}}{\left|\mathrm{a}\right|\left|\mathrm{b}\right|}<0\\ \text{ i.e., }& \frac{{\mathrm{cx}}^2-12+6\mathrm{cx}}{\sqrt{{\mathrm{c}}^2{\mathrm{x}}^2+36+9}\sqrt{{\mathrm{x}}^2+4+4{\mathrm{c}}^2{\mathrm{x}}^2}}<0\\ \Rightarrow & {\mathrm{cx}}^2+6\mathrm{cx}-12<0\end{array}$

Now, two cases are possible
Case $\mathrm{I}$    $\mathrm{c}\ne 0$
$\Rightarrow {\mathrm{cx}}^2+6\mathrm{cx}-12$is a equation  which has real solution

$$\begin{gathered} \text{ "iff }\mathrm{a}<0\text{ and }{\mathrm{b}}^2-4\mathrm{ac}<0\text{ " } \\ \begin{array}{l}\text{ i.e., if }\mathrm{c}<0\text{ and }36{\mathrm{c}}^2+48<0\\ \text{ i.e., if }\mathrm{c}<0\text{ and }12\mathrm{c}(3\mathrm{c}+4)<0 \\ \begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{c}+4>0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\because \mathrm{c}>0]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-\frac{4}{3}<\mathrm{c}<0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\dots \dots \text{ (ii) }\end{array}\end{array} \end{gathered}$$
Case $\mathrm{II}$    $\mathrm{c}=0$
 which is an identity
c = 0 satisfy Eq (i)
From Eqs (ii) and (iii), we get $-\frac{4}{3}<\mathrm{c}\le 0$