Find the equation of the cylinder whose generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3$ and whose guiding curve is the ellipse Find the equation of the cylinder whose generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3$ and whose guiding curve is the ellipse $x^2+2y^2=1,z=3.$
My solution goes like this:

The generators are parallel to the line $\frac x1=\frac {-y}{2}=\frac z3.$ So, the equation of the generator is $\frac {x-\alpha}{1}=\frac {y-\beta}{-2}=\frac {z-\gamma}{3},$ where $(\alpha, \beta,\gamma)$ is a point on the cylinder. As we know, from the definition of a cylinder, it is a generator line which follows the trajectory of the guiding curve, just like a remote star. The guiding curve in this case is an ellipse. The cylinder extends infinitely in space and hence the cylinder also intersects in the $xy$ plane and in that plane the cylinder also follows the trajectory of the given ellipse. We can consider this situation alternatively, by the considering the guiding curve as $x^2+2y^2=1,z=0,$ because this will generate the same cylinder as the one in the  case of $x^2+2y^2=1,z=3.$ So, in the $xy$ plane the point $(x,y,z)$ of the generater needs a closer examination. In the $xy$ plane, $z=0$ and hence $\frac {x-\alpha}{1}=\frac {y-\beta}{-2}=\frac {-\gamma}{3},$  due to which, $x=\alpha+\frac{-\gamma}{3}$ and $y=\beta+\frac{2\alpha}{3}.$ Thus, the point $(\alpha+\frac{-\gamma}{3},\beta+\frac{2\alpha}{3},0)$ on the generator satisfies the equation of $x^2+2y^2=1.$ Therefore, $(\alpha+\frac{-\gamma}{3})^2+2(\beta+\frac{2\alpha}{3})^2=1$ is the locus of $(\alpha, \beta,\gamma)$ which is the cylinder. Thus, the equation of the cylinder is  $(\alpha+\frac{-\gamma}{3})^2+2(\beta+\frac{2\alpha}{3})^2=1,$ when written in terms of $x,y,z$ instead of $\alpha, \beta,\gamma$, i.e, $(x+\frac{-z}{3})^2+2(y+\frac{2x}{3})^2=1.$

Is the above solution correct? If not then where is it going wrong?

Your solution is wrong at that precise point: "We can consider this situation alternatively, by the considering the guiding curve as $x^2+2y^2=1,z=0,$ because this will generate the same cylinder as the one in the case of $x^2+2y^2=1,z=3$" because the "generators" are not vertical.
There was a (small) second mistake when you solved $\frac {y-\beta}{-2}=\frac {-\gamma}{3}$: you wrote $y=\beta+\frac{2\alpha}{3}$ instead of $y=\beta+\frac{2\gamma}{3}.$
If you make the necessary corrections, your final equation will be
$$\left(x+\frac{-z}{3}+1\right)^2+2\left(y+\frac{2z}{3}-2\right)^2=1.$$
