In 3 dimensional euclidean space, I am looking for a plane. I have the following information:
Assume without loss of generality that $v$ is a unit vector.
Let $n$ be the unit vector normal to the sought plane, with non-negative $n_y$.
The angle between $n$ and the y axis is equal to the angle between the sought plane and the XZ plane:
$\cos(i) = n\cdot \left[ {\begin{array}{c} 0 \\ 1 \\ 0 \end{array} } \right]\Rightarrow n_y=\cos(i).$
The vectors $n$ and $v$ are orthogonal:
$n\cdot v=0\Rightarrow n_x=\frac{- n_{y} v_{y} - n_{z} v_{z}}{v_{x}}$
The vector $n$ is a unit vector:
$n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = 1\Rightarrow n_{z1,2}=\pm \sqrt{1-n_x^2 - n_y^2}$
Inserting $n_{z1}$ in the solution for $n_x$ yields:
$$\begin{align}n_x&=\frac{- n_{y} v_{y} + v_{z} \sqrt{- n_{x}^{2} - n_{y}^{2} + 1}}{v_{x}}\\\Rightarrow n_{x1,2}&=\frac{- n_{y} v_{x} v_{y} \pm v_{z} \sqrt{- n_{y}^{2} v_{x}^{2} - n_{y}^{2} v_{y}^{2} - n_{y}^{2} v_{z}^{2} + v_{x}^{2} + v_{z}^{2}}}{v_{x}^{2} + v_{z}^{2}}\end{align}$$
Inserting $n_{x1,2}$ into $n_{z1}$ yields two solutions for $n_z$, which shall be referred to as $n_{z11}$ and $n_{z12}.$
Inserting $n_{z2}$ into the solution for $n_x$ yields the same $n_{x1,2}$, which inserted into $n_{z2}$ yield $n_{z21}$ and $n_{z22}$.
In total there are four solutions:
$$\left[ {\begin{array}{c} n_{x1} \\ n_y \\ n_{z11} \end{array} } \right],\left[ {\begin{array}{c} n_{x2} \\ n_y \\ n_{z12} \end{array} } \right],\left[ {\begin{array}{c} n_{x1} \\ n_y \\ n_{z21} \end{array} } \right],\left[ {\begin{array}{c} n_{x2} \\ n_y \\ n_{z22} \end{array} } \right],$$
of which one is hopefully sound.
As for the constraint with regards to the angle $\theta$ and vector $r$: I think now that it is irrelevant. For any angle $\theta$ and vector $v$ in a plane there are an infinite number of vectors with angle $\theta$ relative to $v$ in that plane.
The second condition involving $\theta$ is as the OP said irrelevant.
Let $\mathbf{n}$ be the normal to the plane, then
$ \mathbf{n} \cdot \mathbf{v} = 0$
Let $\mathbf{u_1}, \mathbf{u_2}$ be two unit vectors that are both orthogonal to $\mathbf{v}$ and to each other, then
$ \mathbf{n} = \cos \theta \ \mathbf{u_1} + \sin \theta \ \mathbf{u_2} $
The condition involving angle $i$, implies the angle between the normals to the $XZ$ plane and our plane is $i$, i.e.
$ \cos i = \mathbf{n} \cdot \mathbf{j} $
This equation can have zero, one, or two solutions for $\theta$.
As a numerical example, suppose
$ \mathbf{v} = (1, 2, 2 ) $
Then we can take
$\mathbf{u_1} = (2, -1, 0)/\sqrt{5} $
$\mathbf{u_2} = \dfrac{1}{3} \mathbf{v} \times \mathbf{u_1} = \dfrac{1}{3\sqrt{5}} (2 , 4, -5) $
Suppose the angle $i = 60^\circ $, then
$\cos i = \cos 60^\circ = \dfrac{1}{2} = \cos \theta \bigg(-\dfrac{1}{\sqrt{5}}\bigg) + \sin \theta \bigg( \dfrac{ 4 }{3 \sqrt{5}} \bigg) $
This results in the following two values for $\theta$
$\theta_1 = 1.37881556, \theta_2 = 3.04977931 $
Vector $\vec{v}$ is not just any vector, as it is defined to be in a plane that has angle $i$ with the XZ plane. Therefore knowing $\vec{v}$ effectively fixes $i.$ Indeed, $i=\pi/2-\cos^{-1}(\hat{v}\cdot \hat{n}_1)$ where $n_1=(0,1,0)=\text{ the y-axis.}$ Even though $i,$ the angle of our sought plane with plane XZ is known, that still leaves an infinity of planes which satisfy the condition. For my convenience, I assure that all vectors are converted to unit vectors before calculation.
All of the planes that make angle $i$ with plane XZ will lie along the edge of a cone. Said cone will have the y-axis as its axis with its vertex at the origin. Its equation will be $$x^2 + z^2 = (y\cdot \tan(\pi/2 - i))^2.$$ For reference to this see 1224620.
In the picture, the grid is the XZ plane and the blue plane is "the plane that we seek".
The normal to this plane necessarily must also be a normal to the cone, which will just be the coefficients of the Cartsian equation multiplied times any point C which is on the surface of the cone. But, we do need to carefully select point C.
Since we have built the cone to include vector $\vec{v}$, we can let point C be the end point of the vector. $C=v.$
Here are the coefficients.
$$\begin{align}\text{Coefficient of x^2: }&1\\
\text{Coefficient of y^2: }&-\tan^2\left(\frac{2i-\pi}{2}\right)\\
\text{Coefficient of z^2: }&1
\end{align}$$
and the normal to the plane
$$\hat{n}=\frac{\left(
\begin{array}{ccc}
\hat{v}_x\\
-\tan^2\left(\frac{2i-\pi}{2}\right)\cdot \hat{v}_y\\
\hat{v}_z
\end{array}
\right)}
{\left\Vert \left(
\begin{array}{ccc}
\hat{v}_x\\
-\tan^2\left(\frac{2i-\pi}{2}\right)\cdot \hat{v}_y\\
\hat{v}_z
\end{array}
\right)\right\Vert}
$$
The plane equation:
$$\hat{n}\cdot \left(\begin{array}{ccc}x\\y\\z\end{array} \right)=\hat{n}\cdot C\qquad \text{remember point C=v}$$
We still have to deal with $\theta,$ to get vector $\vec{r}$ in the same plane, but that seems straigtforward. Remember that $\theta$ is given while $\vec{r}$ is not given.
We need an orthogonal basis for the plane. We already have one unit vector in the plane, namely $\hat{v}.$ We also have the unit normal vector $\hat{n}.$ Cross them to get a 2nd basis vector in the plane. $$\hat{u}=\frac{n\times v}{|n\times v|}$$
Now just use a rotation in the plane to find some vector that is $\theta$ degrees away from $v,$ but in the plane. For a reference to this see 1830695
$$r = cos(\theta)\hat{v}+sin(\theta)\hat{u}$$
