Beware, it's not IMO problem!

Let $x\geq y\geq z \\ 1/x \leq 1/y \leq 1/z \\ 1/x^2 \leq 1/y^2 \leq 1/z^2$

$\frac{1}{x^2}+\frac{2}{y^2}+\frac{3}{z^2} \leq \frac{1}{z^2} +\frac{2}{z^2} +\frac{3}{z^2}=\frac{6}{z^2} \\ \implies \frac{6}{z^2} \geq 2/3 \\ \implies z^2 \leq 9 \\ \implies |z| \leq 3$

Here if z is a solution then -z is also a solution

Now putting $|z|=3$ we get:

$\frac{1}{x^2} + \frac{2}{y^2}=2/3-1/3=1/3$

Again applying inequality $x \geq y$ we get $|y| \leq 3$

Now subcases $|y|=1,2,3$ only $y\pm 3$ gives solution $x=\pm 3$

It can be easily checked that other cases subcases give no solution.

So there are $8$ triplets are $(x,y,z)=\pm 3,\pm 3,\pm 3$