<s>[INST] 3-D geometry: line intersecting 2 lines and parallel to plane Find a surface generated by line intersecting lines $$y=a=z$$ and $$x+3z=a=y+z$$ and parallel to plane $$x+y=0$$
I tried to form a line equation which intersects the  given two lines i.e.
$(y-a)+k1(z-a)=0$ and $(x+3z-a)+k2(y+z-a)=0$. But don't know how to use the other (plane) condition. 
 [/INST] Reordering the equation of the line
$$\alpha: y+k_1 z-a k_1-a=0;\;\beta:x+k_2 y+(k_2+3) z -ak_2-a=0$$
The normal vector of plane $\alpha$ is $n_1=(0,1,k_1)$ and the normal vector of $\beta$ is $n_2=(1,k_2,k_2+3)$
the direction vector of the line is the cross product $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$
The line $(y-a)+k_1(z-a)=0;\;(x+3z-a)+k_2(y+z-a)=0$ is parallel to the plane $x+y=0$ if its direction vector is perpendicular to the normal vector of the plane $x+y=0$ which is $n=(1,1,0)$
$m\cdot n=0\to (3+k_2-k_1 k_2,k_1,-1)\cdot (1,1,0)=3 + k_1 + k_2 - k_1 k_2$
the condition is that $3 + k_1 + k_2 - k_1 k_2=0$
$k_2=\dfrac{k_1+3}{k_1-1}$
$\alpha:y+k_1 z-a-a k_1=0$
$\beta:(-1+k_1) x+(3+k_1) y+(3+3 (-1+k_1)+k_1) z-a (-1+k_1)-a (3+k_1)=0$
The direction of the line is $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$ where $k_2=\dfrac{k_1+3}{k_1-1}$
$m=(k_1^2-k_1,k_1-k_1^2,k_1-1)$
a point on the line is $P\left(-6 a,0,\dfrac{3 a}{2}\right)$
a parametric equation of the line is
$(x,y,x)=P+tm$ that is
$x=-k_1 t+k_1^2 t-6a,y=k_1 t-k_1^2 t,z=\dfrac{3 a}{2}-t+k_1 t$
Note that from the first two equations $x=-y-6a$
Furthermore from the three equation we get
$k_1= -\dfrac{4 y}{x+y+4 z},\;t= -\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}$
substitute in the third equation $z=\dfrac{3 a}{2}-t+k_1 t$
$z=\dfrac{3 a}{2}+\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}+\dfrac{y (x+y+4 z)}{x+5 y+4 z}$
simplify and get the result.
The requested surface is the plane $x + y+6a=0$
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