I got a plane with equation: $P(x,y)=\frac{16100+47y}{52}$
A line (parametrized equation) which lies on the plane $P$: $$
\left\{
\begin{array}{c}
x=t \\
y=620 \\
z=870
\end{array}
\right.
$$
And a point which belongs to the line: $O(-5,620,870)$
I need to find an equation of the line that belongs to the specified plane $P$ and crosses specified line at the specified point $O$ making an angle of $45^\text{o}$ with the line.
Here's the respective image from Mathcad: 
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 $$\mathbf n \cdot \mathbf r = 16100$$ 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 $$\mathbf n \cdot O = (0)(-5) + (-47)(620) + (52)(870) = -29140 + 45240 = 16100$$ You can plug this back into the equation for $P$ as: $$\mathbf n \cdot \mathbf r = \mathbf n \cdot O$$ and rearrange to get $$\mathbf n \cdot (\mathbf r - O) = 0$$ 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 $\mathbf n$ is perpendicular to all directions along the plane $P$. Indeed, $P$ is the unique plane which is perpendicular to $\mathbf n$ and passes through the point $O$.
The line $\ell$ 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 $\mathbf n \cdot (t\mathbf i) = t(\mathbf n \cdot \mathbf i) = t(0) = 0$, which shows that the line does lie in $P$.
To find the $45^\circ$ lines, first find a line in the plane perpendicular to $\ell$. Fortunately the cross product provides what we need. Set $$\mathbf m = \mathbf n \times \mathbf i = (0, 52, 47)$$ Then $\mathbf m$ is perpendicular to both $\mathbf n$ and $\mathbf i$: $$\mathbf n\cdot \mathbf m = (0)(0) + (-47)(52) + (52)(47) = 0\\\mathbf i\cdot \mathbf m = (1)(0) + (0)(52) + (0)(47) = 0$$ So the line $\ell^\perp$ parametrized by $O+t\mathbf m$ lies in the plane $P$ and is perpendicular to $\ell$.
$\mathbf i$ is a unit vector: $\|\mathbf i\| = \sqrt{1^2 + 0^2 +0^2} = 1$. But $\mathbf m$ is not: $\|\mathbf m\| = \sqrt{0^2 + 52^2 + 47^2} = \sqrt{4913} \approx 70.1$. We need unit vectors for the next step, so divide $\mathbf m$ by its length: $$\mathbf{\hat m} = \frac1{\|\mathbf m\|}\mathbf m = \left(0,\frac{52}{\sqrt{4913}}, \frac{47}{\sqrt{4913}}\right)$$
Finally we can rotate $\mathbf i$ by $45^\circ$ towards $\mathbf{\hat m}$ or towards $-\mathbf{\hat m}$ to get vectors pointing along the two $45^\circ$ lines. For most angles you would use trig for this, but $45^\circ$ is particularly easy: $$\mathbf u_1 = \frac1{\sqrt 2}\left(\mathbf i + \mathbf{\hat m}\right) = \left(\frac 1{\sqrt 2}, \frac{52}{\sqrt{9826}},\frac{47}{\sqrt{9826}}\right)\\ \mathbf u_2 = \frac1{\sqrt 2}\left(\mathbf i - \mathbf{\hat m}\right) = \left(\frac 1{\sqrt 2}, -\frac{52}{\sqrt{9826}},-\frac{47}{\sqrt{9826}}\right)$$
The parametric expressions for the two lines are $O + t\mathbf u_1$ and $O + t\mathbf u_2$.