The question is to find the values of $a$ for which $X_a = \{(x,y,z)\in R^3 | x^2+y^2+z^2 = a\}$ and $Y=\{(x,y,z)\in R^3 | x+y^2+2z = 1\}$ intersect transversally in $R^3$.
If $Y$ had $x^2$ instead of $x$, this would have been much easier. To find $X_a \cap Y$, other than getting $x^2+z^2-x-2z = a-1$, I am not sure how to simply further to get a condition on $a$. Any hint is appreciated. Thanks.
Following the hints, $T(X_a) = \{(u,v,w)| xu+yv+zw=0\}$ and $T(Y) = \{(u,v,w)|xu+2yv+2zw=0\}$. These two tangent spaces are the same when $x=0$ which means $y^2+z^2 = a$ and $y^2+2z=1$. Thus $(z-1)^2=a$ which means $a\geq 0$ which means $X_a$ and $Y$ intersect nontransversally when $a\geq 0$ and intersect transversally when $a<0$. Does this look right?
From calculus 3, the gradient of $f(x,y,z)=x^2+y^2+z^2$ evaluated at any $(x,y,z)$ satisfying $f(x,y,z)=a$ is a vector that is perpendicular to the surface $f(x,y,z)=a$. Similarly for $g(x,y,z)=x+y^2+2z$, its gradient, and the surface $g=1$.
Note that $X_a$ and $Y$ are surfaces and, generically, they will be transverse. So it will be easier to find where they are not transverse. If this happens, the tangent space of their intersection will fail to span 3-space. So, their tangent spaces will be the same. this happens whenever the normals are parallel, hence Jason's comment.
This is a long-winded way of saying use the method of lagrange multipliers: $$\nabla f=\lambda\nabla g$$ And, as usual in this situation, careful with your case analysis!