$x^2+2xy-3y^2=0$

Differentiate the above curve,

$2x+2\left[xy\text{'}+y\right]-6yy\text{'}=0$

So, at $\left(2, 2\right), y\text{'}=1$.

So, equation of normal to the curve at $\left(2, 2\right)$ is $x+y=4$.

Thus, perpendicular distance from origin on the normal is

$\left|\frac{0+0-4}{\sqrt{1^2+1^2}}\right|$

$=\left|\frac{4}{\sqrt{2}}\right|=2\sqrt{2}$.