Here I'll give an answer for arbitrary dimension.
For $\vec{x},\vec{w}_1\in \mathbb{R}^n$, what is the Fourier transform along the line $\vec{w}_1\cdot\vec{x} + c = 0$?
Assume $\vec{w}_1$ is normalized.
Using the dirac delta to define the line in $\mathbb{R}^n$ we therefore must calculate,
\begin{align}
\mathcal{F}[\delta(\vec{w}_1\cdot\vec{x} + c)](\vec{\nu}) = \int d\vec{x} \, \delta(\vec{w}_1\cdot\vec{x} + c) \exp{[-2\pi i \vec{x}\cdot\vec{\nu}]},\tag{1}
\end{align}
where $\vec{\nu}$ are the fourier space variables.
Let $\{\vec{w}_2,...,\vec{w}_n\}$ be the orthonormal basis that spans $\mathrm{Null}[\vec{w}_1]$.
You can determine this for any dimension with the Wolfram command NullSpace
, e.g. for 3D.
Define a change of variables by,
\begin{align}
s_1 =& \vec{w}_1\cdot\vec{x} + c\\
s_2 =& \vec{w}_2\cdot\vec{x}\\
&...\\
s_n =& \vec{w}_n\cdot\vec{x}.
\end{align}
Written more precisely,
\begin{align}
\vec{s} =& W \vec{x} + \hat{s}_1 c\\
I =& W W^T. \tag{2}
\end{align}
The second line comes from the orthonmality of the basis of choosen to span the null space of $\vec{w}_1$.
Note: The determinant of an orthogonal matrix satisfying (2) is $1$, and represents a rotation matrix, and therefore the Jacobian of the transform is $1$.
With these new variables equation (1) becomes,
\begin{align}
\mathcal{F}[\delta(s_1)](\vec{\nu}) = \int d\vec{s} \, \delta(s_1) \exp{[-2\pi i W^T(\vec{s}-\hat{s}_1 c)\cdot\vec{\nu}]}.
\end{align}
Now we carry out the integral over $s_1$ by using $\vec{s} = \sum_{i=1}^n s_i \hat{s}_i$,
\begin{align}
\mathcal{F}[\delta(s_1)](\vec{\nu}) =& \int ds_n ... \int ds_1 \, \delta(s_1)\exp{[-2\pi i W^T(\sum_{i=1}^n s_i \hat{s}_i-\hat{s}_1 c)\cdot\vec{\nu}]}\\
=& \exp{[2\pi i c(W^T\hat{s}_1)\cdot\vec{\nu}]} \int ds_n ... \int ds_2 \, \exp{[-2\pi i \sum_{i=2}^n s_i (W^T\hat{s}_i)\cdot\vec{\nu}]}.
\end{align}
Performing the remaining integrations using the dirac delta definition $\delta(y) = \int dp\, \exp[-2\pi i p y]$, we get,
\begin{align}
\mathcal{F}[\delta(s_1)](\vec{\nu}) =& \exp{[2\pi i c(W^T\hat{s}_1)\cdot\vec{\nu}]} \prod_{i=2}^n \delta(W^T\hat{s}_i\cdot\vec{\nu}).
\end{align}
In order to evaluate $W^T\hat{s}_i$, note the basis vectors in the rotated space are related to the original basis vectors via the same rotation matrix, i.e. $[\hat{s}_1, ..., \hat{s}_n] = W [\hat{x}_1, ..., \hat{x}_n]$.
Therefore $W^T\hat{s}_i = \hat{x}_i = \hat{x}_i (\vec{w}_1\wedge\vec{w}_1 + ... + \vec{w}_n\wedge\vec{w}_n)$.