I know I have the correct value of $x, y, z$ but I get a different value for the volume.
My workings:
$$ f=xyz, g=2x^2+3y^2+4z^2-12, \psi=f+\lambda g $$
$$ \psi_x = yz+4\lambda x=0 $$ $$\psi_y = xz+6\lambda y=0 $$ $$\psi_z = xy+8\lambda z=0$$ $$2x^2+3y^2+4z^2=12$$
Solving the simultaneous equations gives,
$$ x=\sqrt2, y=\frac2{\sqrt3}, z=1$$
I know this bit is correct but when I try to calculate the volume my answer is incorrect
$$f(\sqrt2,\frac2{\sqrt3},1) =\frac{2\sqrt6}{3} $$
but they give the answer as $$ \frac{16\sqrt6}{3} $$
which is clearly 8 times my answer
I did a similar question and once again I got the correct x,y,z but the volume was 8 times my answer, but I cant see why they are multiplying by 8
The formula you have for the volume ($V=xyz$) is the volume of the portion of the box in the first octant. But there are 8 octants, each symmetric to the first, so the volume should really be calculated as
$$V=8xyz$$
Updating your solution with the new objective function, it is clear that $x,y,z$ do not change--only the lagrange multiplier $\lambda$ will be 8 times its previous value.
Plugging in the $x,y,z$ you found, we get that the volume is $\frac{16\sqrt{6}}{3}$, as the given solutions indicate.
To help with the intuition surrounding this problem, we can reduce it to the 2D case. Consider the (poorly drawn) image below, and let $(x,y)$ be the coordinates of the upper right corner of the red box.
Then you have that the area of the red box is $A=xy$, but the acutal area of the insctibed rectangle is the orange box, given by $A=4xy$, doubling once for each dimension.
In the 3D case, there are 3 dimensions in which the volume needs to be doubled, but the concept is the same.