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. 

Equation of the line L1 in symmetric form is $$\frac{x}{1} = \frac{y - a}{0} = \frac{z - a}{0} \tag1$$
Equation of the Line L2 in symmetric form is $$\frac{x -a}{-3} = \frac{y - a}{-1} = \frac{z}{1} \tag2$$
Let the required line 'L' intersect L1 at P and L2 at Q.
Coordinates of P are $$(r,a,a) \tag3$$
Coordinates of Q are $$(-3s+a, -s+a, s) \tag4$$
Equation of L is  $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} = \frac{z-a}{s-a} \tag5$$
Given that this line L is parallel to the plane $x+y=0 \tag6$
Hence, its normal will be perpendicular to L
So $$(-3s-r+a)+(-s)=0 \tag7$$
or $$r=a-4s \tag8$$
From equation 5 we have $$\frac{y-a}{z-a} = \frac{-s}{s-a}$$
Solving this for s, we will have $$s=\frac{a(y-a)}{y+z-2a} \tag9$$
From equation 5, we also have $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} \tag10$$
Substituting the values of r and s obtained from equations 8 and 9, we get the required surface as
$$y^2 + xy + yz + zx - 2ax -2az = 0$$
