Given equations for two planes,
$$a_1x + b_1y + c_1 z + d_1 = 0 $$
$$a_2x + b_2y + c_2 z + d_2 = 0 $$
...and a reference point $(x_o,y_o,z_o) $ that exists on the line of intersection of these planes...
What are the equations for the two points $(x_p,y_p,z_p)_+ $ and $(x_p,y_p,z_p)_{-}$ that are each distance $D$ from the reference point on the line?
On
Here's a somewhat more 'linear-algebra' solution: note that your line will have an equation $\vec{p}=\vec{p_0}+t\vec{v}$, where $\vec{p_0}$ is an arbitrary point on the line. Now, since this line is an intersection of two planes $P_1=\{\vec{p}:\vec{p}\cdot\vec{n_1}=d_1\}$ and $P_2 = \{\vec{p}:\vec{p}\cdot\vec{n_2}=d_2\}$, then $\vec{v}$ (that is, the ray along the line) will lie in both planes — in other words, it will be orthogonal to $\vec{n_1}$ and $\vec{n_2}$. But the canonical vector orthogonal to $\vec{n_1}$ and $\vec{n_2}$ is $\vec{v}=\vec{n_1}\times\vec{n_2}$, the cross product; normalizing this vector to a unit vector $\hat{v}$ and then taking $\vec{p_\pm}=\vec{p_0}\pm D\hat{v}$ will give the two points you're looking for.
I solved this using the distance formula for two points:
$$D = \sqrt{(x_p - x_o)^2 + (y_p - y_o)^2 + (z_p - z_o)^2} $$
Starting off with the two plane equations, I set $z=t$ in each of them, and then between those two equations solved for $x$ and $y$ purely in terms of $t$.
Then I plugged these x(t) and y(t) values in terms of $t$ into the above distance equation, to solve for the two possible values of $t$. Then I plugged these values back into the values of $x,y,z$.