I got a plane with equation: P ( x , y ) = 16100 +

Step 1
That is not the equation of a plane. It is not really an equation at all. It is the definition of the "P(x,y)" function. An equation forces the variables in it into some relationship. If you don't count the function P as a variable, this does no such thing. From the rest of your post, it is evident that the equation of the plane is
$z=\frac{16100+47y}{52}$
which forces a relationship between the y and z coordinates of the points on the plane. (Since x does not appear, it can still take on any value. But specifying a value for either y or z forces a value on the other by the equation.)
Solving your problem is most easily done using vector notation. You can rewrite the equation of the plane P as
$-<!--\; -\; -->47y+52z=16100$
Or
$\mathbf{n}\cdot <!--\; \cdot \; -->\mathbf{r}=16100$
where $\mathbf{r}=(x,y,z)$ is the position vector, representing points in space as vectors from the origin, and $\mathbf{n}=(0,-<!--\; -\; -->47,52)$ is a vector. The points of P are exactly those $\mathbf{r}$ for which the equation holds.
$O=(-5,620,870)$ can be thought of as a vector now. We can verify that it is a point on the plane by calculating
$\mathbf{n}\cdot <!--\; \cdot \; -->O=(0)(-<!--\; -\; -->5)+(-<!--\; -\; -->47)(620)+(52)(870)=-<!--\; -\; -->29140+45240=16100$
You can plug this back into the equation for P as:
$\mathbf{n}\cdot <!--\; \cdot \; -->\mathbf{r}=\mathbf{n}\cdot <!--\; \cdot \; -->O$
and rearrange to get
$\mathbf{n}\cdot <!--\; \cdot \; -->(\mathbf{r}-<!--\; -\; -->O)=0$
as an alternate form of the equation of the plane. Since $\mathbf{r}$ and O are points on the plane, their difference is a direction along the plane. So this says that the vector n is perpendicular to all directions along the plane P. Indeed, P is the unique plane which is perpendicular to n and passes through the point O.
The line $\mathrm{\ell <!--\; \ell \; -->}$ is defined parametrically by the expression
$O+t\mathbf{i}$
where $\mathbf{i}=(1,0,0)$ For every value of t, the point $O+t\mathbf{i}$ lies on the line, and for every point on the line, there is some t for which that point is $O+t\mathbf{i}$ . Note that if $\mathbf{r}=O+t\mathbf{i}$ , then $\mathbf{r}-<!--\; -\; -->O=t\mathbf{i}$ and $\mathbf{n}\cdot <!--\; \cdot \; -->(t\mathbf{i})=t(\mathbf{n}\cdot <!--\; \cdot \; -->\mathbf{i})=t(0)=0$ , which shows that the line does lie in P.
To find the ${45}^{\circ <!--\; \circ \; -->}$ lines, first find a line in the plane perpendicular to $\mathrm{\ell <!--\; \ell \; -->}$ . Fortunately the cross product provides what we need. Set
$\mathbf{m}=\mathbf{n}\times <!--\; \times \; -->\mathbf{i}=(0,52,47)$
Then m is perpendicular to both n and i:
$$\begin{gathered} \mathbf{n}\cdot <!--\; \cdot \; -->\mathbf{m}=(0)(0)+(-<!--\; -\; -->47)(52)+(52)(47)=0 \\ \mathbf{i}\cdot <!--\; \cdot \; -->\mathbf{m}=(1)(0)+(0)(52)+(0)(47)=0 \end{gathered}$$
So the line ${\mathrm{\ell <!--\; \ell \; -->}}^{\perp <!--\; \perp \; -->}$ parametrized by $O+t\mathbf{m}$ lies in the plane P and is perpendicular to $\mathrm{\ell <!--\; \ell \; -->}$ .
i is a unit vector: $\Vert <!--\; \Vert \; -->\mathbf{i}\Vert <!--\; \Vert \; -->=\sqrt{1^2+0^2+0^2}=1$ . But m is not: $\Vert <!--\; \Vert \; -->\mathbf{m}\Vert <!--\; \Vert \; -->=\sqrt{0^2+{52}^2+{47}^2}=\sqrt{4913}\approx <!--\; \approx \; -->70.1$ . We need unit vectors for the next step, so divide m by its length:
$\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{m}}=\frac{1}{\Vert <!--\; \Vert \; -->\mathbf{m}\Vert <!--\; \Vert \; -->}\mathbf{m}=(0,\frac{52}{\sqrt{4913}},\frac{47}{\sqrt{4913}})$
Finally we can rotate i by ${45}^{\circ <!--\; \circ \; -->}$ towards $\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{m}}$ or towards $-<!--\; -\; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{m}}$ to get vectors pointing along the two ${45}^{\circ <!--\; \circ \; -->}$ lines. For most angles you would use trig for this, but ${45}^{\circ <!--\; \circ \; -->}$ is particularly easy:
$$\begin{gathered} {\mathbf{u}}_1=\frac{1}{\sqrt{2}}(\mathbf{i}+\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{m}})=(\frac{1}{\sqrt{2}},\frac{52}{\sqrt{9826}},\frac{47}{\sqrt{9826}}) \\ {\mathbf{u}}_2=\frac{1}{\sqrt{2}}(\mathbf{i}-<!--\; -\; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{m}})=(\frac{1}{\sqrt{2}},-<!--\; -\; -->\frac{52}{\sqrt{9826}},-<!--\; -\; -->\frac{47}{\sqrt{9826}}) \end{gathered}$$
The parametric expressions for the two lines are $O+t{\mathbf{u}}_1$ and $O+t{\mathbf{u}}_2$