Let the given vectors $\hat{\mathrm{i}}-2 \mathrm{x} \hat{\mathrm{j}}-3 \mathrm{y} \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+3 \mathrm{x} \hat{\mathrm{j}}+2 \mathrm{y} \widehat{\mathrm{k}}$.be $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$
respectively, and $\theta$ be the angle between
them, so, $\cos \theta=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}|}$
These are orthogonal to each other, $\theta=\pi / 2$ so, $\overrightarrow{\mathrm{A} \cdot \overrightarrow{\mathrm{B}}}=0$
$\Rightarrow 1-(2 x)(3 x)-(3 y)(2 y)=0$
$\Rightarrow 1-6 x^{2}-6 y^{2}=0$
$\Rightarrow x^{2}+y^{2}=\frac{1}{6}$
This equation represents an equation of a circle which is the locus of the point $(\mathrm{x}, \mathrm{y})$