We know, the sum of the distances from any point $P(h,k)$ on the ellipse to those two foci is constant and equal to the major axis $(2a)$.
As $(2,\sqrt2)$ lies on the ellipse, $2a=\sqrt{(3-2)^2+(0-\sqrt 2)^2}+\sqrt{(-3-2)^2+(0-\sqrt 2)^2}=\sqrt 3+3\sqrt3=4\sqrt3$
So, $\sqrt{(h-3)^2+(k-0)^2}+\sqrt{\{h-(-3)\}^2+(k-0)^2}=2a=4\sqrt 3$
or ,$\sqrt{(h-3)^2+(k-0)^2}=4\sqrt 3-\sqrt{\{h-(-3)\}^2+(k-0)^2}$
On squaring, $(h-3)^2+(k-0)^2=48+(h+3)^2+k^2-8\sqrt3\sqrt{(h+3)^2+k^2}$
or, $8\sqrt3\sqrt{(h+3)^2+k^2}=48+12h$
On squaring and simplification, $h^2+4k^2=12\space or \space \frac {h^2}{12}+\frac{k^2}3=1$
So, the locus of $P(h,k)$ is $\frac {x^2}{12}+\frac{y^2}3=1$
Alternatively,
$a=2\sqrt 3$.
Here $ae=3,e=\frac3{2\sqrt3}=\frac{\sqrt3}2$
If $2b$ is the minor axis,$b^2=a^2(1-e^2)=(2\sqrt3)^2\left(1-\frac3 4\right)=3$
We know, the midpoint of the segment connecting the foci is the centre of the ellipse, so here the centre is $\frac{3-3}2,\frac{0+0}2$ i.e., $(0,0)$
Again, the major axis is the segment that contains both foci.
Here the equation of the major axis is $\frac{y-0}{x-3}=\frac{0-0}{\{3-(-3)\}}$ i.e, $y=0$ the X axis.
So, the required equation of the ellipse is $\frac{(x-0)^2}{(2\sqrt3)^2}+\frac{(y-0)^2}{3}=1 $