In a triangle $ABC,$ right angled at the vertex $A,$ if the position vectors of $A,$ Band Care respectively 

$3\widehat{i}+\widehat{j}-\widehat{k},-\widehat{i}+3\widehat{j}+p\widehat{k}\text{ and }5\widehat{i}+q\widehat{j}-4\widehat{k}$ , then the point 

$(p, q)$ lies on a line 

We have, $\overrightarrow{AB}=-4\widehat{i}+2\widehat{j}+(p+1)\widehat{k}$ 

$\overrightarrow{AC}=2\widehat{i}+(q-1)\widehat{j}-3\widehat{k}$

$ABC$ is a right angled triangle, right angle at $A.$

 $\begin{array}{l}\therefore \overrightarrow{AB}\perp \overrightarrow{AC}\Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AC}=0\\ \Rightarrow -8+2(q-1)-3(p+1)=0\Rightarrow 3p-2q+13=0\\ \therefore (p,q)\text{ lies on the line }3x-2y+13=0\end{array}$

Now, slope of Iine = $\frac{3}{2}$ 

$\therefore$ The point $(p, q)$ lies on a line making acute angle with the positive direction of $x$-axis.