Surfaces and revolution How can i determine the equation of a surface that is obtained from the revolution of the line r around the line s
Given line r
$$
\left\{ 
\begin{array}{c}
2x-y-3=0 \\ 
x+z-2=0 \\
\end{array}
\right. 
$$
And line s : 
$$
\left\{ 
\begin{array}{c}
x-y-2=0 \\ 
z=1 \\ 
\end{array}
\right. 
$$
Is there a specific method that i can use it to solve questions like this ? 
The parametric equation of line r
$$
\left\{ 
\begin{array}{c}
X=2-t \\ 
Y=1-2t \\ 
Z=t
\end{array}
\right. 
$$
For s : 
$$
\left\{ 
\begin{array}{c}
X=2-t\\ 
Y=0-t \\ 
Z=1
\end{array}
\right. 
$$
And if there is a method can i use it also to obtain a surface that generated from a rotation of a line r for example around z-axis or x-axis ?
I have an exam soon :(
 A: We are here in a special case : lines $r$ and $s$ meet at point $C:=(1,-1,1)$.
Therefore the generated surface is a cone.
Considering your parametric representation of line $s$, i.e.,
$$
\left\{ 
\begin{array}{c}
x=2-t\\
y=0-t \\
z=1
\end{array}
\right., 
$$ 
it is clear that $\vec{d}:=(-1,-1,0)$ is a directing vector of this line.
The cone can be described as the set of points $M(x,y,z)$ such that the following absolute value is a constant (this constant will be explicitated afterwards):
$$\left|\dfrac{\vec{CM}}{\|\vec{CM}\|}.\vec{d}\right|=k \ \ \iff \ \ (\vec{CM}.\vec{d})^2=k^2 \|\vec{CM}\|^2 \tag{1}$$
(the dot is a dot product). (1) is equivalent to :
$$((x-1)(-1)+(y+1)(-1)+(z-1)(0))^2=k^2((x-1)^2+(y+1)^2+(z-1)^2)\tag{2}$$
which is the looked for equation of the cone.
But what is the value of $k$ ? It is obtained by taking a particular instance of $M$, call it $M_0$, belonging to line $r$. If we take for example $M_0:=(2,1,0)$, we get :
$$k=\left|\dfrac{\vec{CM_0}}{\|\vec{CM_0}\|}.\vec{d}\right|=\dfrac{3}{\sqrt{6}}.$$
It remains to expand (1)...
A: A very general approach to finding an implicit equation for this surface is to parameterize it and eliminate the parameters. Working with your parameterizations $\mathbf r(t)$ and $\mathbf s(t)$ of the two lines, we let $\mathbf v = \mathbf r(t) - \mathbf s(0)$ and set $\mathbf k$ to the normalized direction vector of $\mathbf s$. By Rodrigues’ rotation formula, $\mathbf r(t)$ rotated about $\mathbf s$ through an angle $\theta$ is $$\mathbf s(0)+\mathbf v \cos\theta+(\mathbf k\times\mathbf v)\sin\theta+\mathbf k(\mathbf k\cdot\mathbf v)(1-\cos\theta),$$ which expands to $$\begin{align} x &= \frac12\left(5-3t+(t-1)\cos\theta+\sqrt2(t-1)\sin\theta\right) \\ y &= \frac12\left(1-3t-(t-1)\cos\theta-\sqrt2(t-1)\sin\theta\right) \\ z &= 1+(t-1)\cos\theta-\frac1{\sqrt2}(t-1)\sin\theta.\end{align}$$ Adding twice the second equation to the third produces $2y-z=2-3t-\frac3{\sqrt2}(t-1)\sin\theta$, from which $$\sin\theta = -{\sqrt2(2y+z+3t-2)\over3(t-1)}.$$ Similarly, we get $$\cos\theta = {2x+2z+3t-7\over3(t-1)}$$ from the first and last equations, whereupon $$2(2y+z+3t-2)^2+(2x+2z+3t-7)^2 = (3(t-1))^2.$$ Finally, adding the first two equations produces $x+y=3(1-t)$, which we can substitute back (and change signs to make it prettier) to get $$2(x-y-z-1)^2+(x-y+2z-4)^2 = (x+y)^2.$$ Observe that the parenthesized expressions on the left-hand side describe a pair of perpendicular planes that contain $\mathbf s$, while the parenthesized expression on the right is the plane perpendicular to $\mathbf s$ that passes through the vertex of the cone. This suggests a perhaps easier way to arrive at an equivalent equation.  
We know that the two lines intersect at $(1,-1,1)$, so the surface is actually a cone—a degenerate hyperboloid. If we choose a coordinate system with origin at the center of the hyperboloid and $Z$-axis aligned with $\mathbf s$, the canonical Cartesian equation in that coordinate system has the form $$X^2+Y^2=\alpha Z^2.$$ Transforming back to the original coordinate system can be accomplished by substituting suitably normalized expressions for the coordinate planes into this equation.  
The $X$-$Y$ plane here has equation $x+y=0$, and for the other two coordinate planes we can choose any convenient pair of orthogonal planes that include $\mathbf s$. The two given defining planes of $\mathbf s$ happen to be orthogonal, so we can use their equations. The cone equation is therefore, after normalizing these plane equations, $$\frac12(x-y-2)^2+(z-1)^2=\frac\alpha2(x+y)^2.$$ Substituting another known point on $\mathbf r$ into this and solving for $\alpha$, we arrive at $$3(x-y-2)^2+6(z-1)^2=(x+y)^2.$$ I’ll leave it to you to verify that this is equivalent to the equation arrived at above by the eliminating parameters.  
If the hyperboloid were not degenerate, we could do something similar. Its canonical equation is $${X^2+Y^2\over a^2}-{Z^2\over c^2}=1.$$ The origin of the coordinate system is the center of the hyperboloid, which is the closest point on $\mathbf s$ to $\mathbf r$. The parameter $a$ is the distance between the two lines, and $c$ can be found by substituting some other known point on $\mathbf r$ besides the closest one to $\mathbf s$ into this equation. In this coordinate system, the rulings of the hyperboloid take the form $(a\cos\alpha,a\sin\alpha,0)+t(-a\sin\alpha,a\cos\alpha,\pm c)$, so another way to find $c$ might be to map $\mathbf r(t)$ into this coordinate system, but I think that’s going to require more work.
