For all real $x$, the vectors $\vec{a}=cx\hat{i}-6\hat{j}+3\hat{k}$ and $\vec{b}=x\hat{i}+2\hat{j}+2cx\hat{k}$ make an obtuse angle with each other. Then the value of $c$ must satisfy:

For an obtuse angle $\theta$, $\cos \theta <0$, which implies $\vec{a}\cdot \vec{b}<0$.$\vec{a}\cdot \vec{b}=(cx)(x)+(-6)(2)+(3)(2cx)<0$$c{x}^2+6cx-12<0$ For this quadratic in $x$ to be negative for all real $x$:The leading coefficient $c$ must be negative: $c<0$.The discriminant $D$ must be less than 0:$D=(6c{)}^2-4(c)(-12)<0$$36c^2+48c<0⟹12c(3c+4)<0$ The roots are $c=0$ and $c=-4/3$. Since $12c(3c+4)<0$, $c$ lies between $-4/3$ and $0$.Condition: $-4/3<c<0$.