Let $L$ be the line of intersection of the planes $~x+y=0~$ and $~y+z=0$.

[b](a)[/b] Write the vector equation of $L$, i.e. find $(a,b,c)$ and $(p,q,r)$ such that $$L=\{(a,b,c)+\lambda(p,q,r)~~\vert~\lambda\in\mathbb{R}\}$$
[b](b)[/b] Find the equation of a plane obtained by $x+y=0$ about $L$ by $45^{\circ}$.
1. **Finding the vector equation of the line \( L \):**

   Let \(\mathcal{P}_1\) be the plane \(x + y = 0\) and \(\mathcal{P}_2\) be the plane \(y + z = 0\). The line \(L\) is the intersection of these two planes.

   - The origin \(O(0,0,0)\) lies on both planes, hence it lies on the line \(L\). Therefore, we can take \((a,b,c) = (0,0,0)\).
   - To find the direction vector \((p,q,r)\) of the line \(L\), we need to find a vector that lies in both planes.

     From the plane \(\mathcal{P}_1\), we have:
     \[
     x + y = 0 \implies x = -y
     \]

     From the plane \(\mathcal{P}_2\), we have:
     \[
     y + z = 0 \implies z = -y
     \]

     Therefore, a point \((x,y,z)\) on the line \(L\) can be written as:
     \[
     (x, y, z) = (-y, y, -y) = y(-1, 1, -1)
     \]

     Letting \(\lambda = y\), we get the direction vector \((p,q,r) = (-1, 1, -1)\).

     Thus, the vector equation of the line \(L\) is:
     \[
     L = \{(0,0,0) + \lambda(-1,1,-1) \mid \lambda \in \mathbb{R}\}
     \]

2. **Finding the equation of a plane obtained by rotating \(x + y = 0\) about \(L\) by \(45^\circ\):**

   Let \(\mathcal{P}_3\) be the plane \(a_3x + b_3y + c_3z = 0\) such that \(\mathcal{P}_1 \cap \mathcal{P}_3 = L\) and the dihedral angle between \(\mathcal{P}_1\) and \(\mathcal{P}_3\) is \(45^\circ\).

   - The normal vector to \(\mathcal{P}_1\) is \(n_1 = (1, 1, 0)\).
   - The normal vector to \(\mathcal{P}_3\) is \(n_3 = (a_3, b_3, c_3)\).

   The dihedral angle \(\alpha\) between \(\mathcal{P}_1\) and \(\mathcal{P}_3\) is the angle between \(n_1\) and \(n_3\):
   \[
   \cos \alpha = \frac{\sqrt{2}}{2} = \frac{|a_3 + b_3|}{\sqrt{2} \cdot \sqrt{a_3^2 + b_3^2 + c_3^2}}
   \]

   Since \(n_3 \perp L\), we have:
   \[
   (a_3, b_3, c_3) \cdot (-1, 1, -1) = 0 \implies -a_3 + b_3 - c_3 = 0 \implies a_3 - b_3 + c_3 = 0
   \]

   Let \(c_3 = k a_3\). Substituting \(c_3 = k a_3\) into the equation, we get:
   \[
   a_3 - b_3 + k a_3 = 0 \implies a_3(1 + k) = b_3 \implies b_3 = a_3(1 + k)
   \]

   Substituting \(b_3 = a_3(1 + k)\) into the cosine equation, we get:
   \[
   \frac{|a_3 + a_3(1 + k)|}{\sqrt{2} \cdot \sqrt{a_3^2 + a_3^2(1 + k)^2 + (k a_3)^2}} = \frac{\sqrt{2}}{2}
   \]

   Simplifying, we get:
   \[
   \frac{|a_3(2 + k)|}{\sqrt{2} \cdot \sqrt{a_3^2(2 + 2k + k^2)}} = \frac{\sqrt{2}}{2}
   \]

   \[
   \frac{|2 + k|}{\sqrt{2 + 2k + k^2}} = \frac{1}{2}
   \]

   Squaring both sides, we get:
   \[
   \frac{(2 + k)^2}{2 + 2k + k^2} = \frac{1}{4}
   \]

   \[
   4(2 + k)^2 = 2 + 2k + k^2
   \]

   \[
   4(4 + 4k + k^2) = 2 + 2k + k^2
   \]

   \[
   16 + 16k + 4k^2 = 2 + 2k + k^2
   \]

   \[
   3k^2 + 14k + 14 = 0
   \]

   Solving this quadratic equation, we get:
   \[
   k = \frac{-14 \pm \sqrt{196 - 168}}{6} = \frac{-14 \pm \sqrt{28}}{6} = \frac{-14 \pm 2\sqrt{7}}{6} = \frac{-7 \pm \sqrt{7}}{3}
   \]

   Therefore, the possible values of \(k\) are:
   \[
   k = \frac{-7 + \sqrt{7}}{3} \quad \text{or} \quad k = \frac{-7 - \sqrt{7}}{3}
   \]

   Substituting these values back, we get the possible planes \(\mathcal{P}_3\):
   \[
   x + (2 + \sqrt{3})y + (1 + \sqrt{3})z = 0 \quad \text{or} \quad x + (2 - \sqrt{3})y + (1 - \sqrt{3})z = 0
   \]

The final answer is \( \boxed{ x + (2 + \sqrt{3})y + (1 + \sqrt{3})z = 0 } \) or \(x + (2 - \sqrt{3})y + (1 - \sqrt{3})z = 0\)