A problem regarding to hyperboloid of one sheet I am confused with following question.
There is a fact that a hyperboloid of one sheet is a doubly ruled suface containing two  groups of lines on its surface.
The problem is to prove that every plane which contains a line in one group also contains a line in the other group.
 A: Your setup is invariant under affine transformations. So without loss of generality, assume the hyperboloid to be $x^2+y^2-z^2=1$ and the line to be $\{(1,t,t)\mid t\in\mathbb R\}$. This line has direction vector $(0,1,1)$ which is orthogonal to both $(1,0,0)$ and $(0,1,-1)$ so you can describe the family of planes through this line as
$$ax+by-bz=a$$
For $b=0$ this is the plane $x=1$ which also contains the line $\{(1,t,-t)\mid t\in\mathbb R\}$ satisfying your claim.
So the more interesting case is $b\neq 0$ for which we can assume $b=1$ since the equation is homogeneous so it can always be scaled to this. In this case, if this plane intersects the hyperboloid in a line from the second family, that second line will pass through the plane $z=0$ at some point other than $(1,0,0)$. With $z=0$ and $b=1$ the equation of the plane yields
$$y = a-ax$$
and plug this into the hyperboloid to obtain
$$x^2+(a-ax)^2=1\qquad\Rightarrow\quad
x\in\left\{1,\frac{a^2-1}{a^2+1}\right\}$$
The first solution is the point where the line from the first family intersects the $z=0$ plane. So let's concentrate on the second, computing $y$ from $x$ as above:
$$x=\frac{a^2-1}{a^2+1}\qquad y=\frac{2a}{a^2+1}$$
Notice how this resembles the tangent half-angle formula. Now the assumption is that rotating $\{(1,t,-t)\mid t\in\mathbb R\}$ around the $z$ axis to this point should yield another line lying fully in the plane. Let's try that:
$$\frac{1}{a^2+1}
\begin{pmatrix}a^2-1&-2a&0\\2a&a^2-1&0\\0&0&a^2+1\end{pmatrix}
\begin{pmatrix}1\\t\\-t\end{pmatrix}=
\begin{pmatrix}(a^2-2ta-1)/(a^2+1)\\(ta^2+2a-t)/(a^2+1)\\-t\end{pmatrix}$$
This is indeed a point from the second family of lines. It satsifies both the equation of the hyperboloid (which it must thanks to the rotation) and of the plane $ax+y-z=a$ (which is the relevant thing here). The latter in more detail:
\begin{align*}
a\frac{a^2-2ta-1}{a^2+1}+\frac{ta^2+2a-t}{a^2+1}+t&=
\frac{a^3-2ta^2-a+ta^2+2a-t+ta^2+t}{a^2+1}\\=
\frac{a^3+a}{a^2+1}&=a
\end{align*}
So yes, there is a line from the second family in this specific plane, which is generic enough to cover all possible planes containing one of the lines for all possible one-sheeted hyperboloids.
