Moment around a point in a frame

Question

Note: Both diagrams are 2d(they might look 3d because I drew them manually.) I have a prismatic frame that looks like this:
I made a section on the half of the first pillar(the red line represents the section), to study the internal forces in it(Normal force, shear force and moment.) Here is a free body diagram at this section:

A is a fixed support so I calculated its reactions as follows:

• $X_A= 100 KN$
• $Y_A= 200 KN$
• $M_A= 300 KN$

Note: $AB=3m$
I have also calculated the normal and the shear force at B, but I have a problem calculating the moment at B and here is why:
now let's apply the equilibrium equation of the moments at B, as:

$$\Sigma M_B=0$$

As you see there are the following moments:

• $X_A\times 3$
• $M_B$
• and at last $M_A$

I have done it like this:
$$ 3X_A + M_B - M_A = 0 $$ I just need to know what to do with $M_A$ should I multiply it with the distance $AB$ or just leave it.(I tried both cases and I think both are wrong). now can somebody till me what to do with the $M_A$?

Edit:

I guess the diagrams are somehow misleading,(I am not good at drawing!). The axis of moment B is the straight line perpendicular to the plane and passing through B and the same for A. Now please look at the first picture: imagine there is a force just above the red section, this force is causing the bending moment in B and the reaction moment in A. I want to calculate the internal moment B, and that's why I need the sum of moments at B.

Answer 1

Edit with respect to correction/edit in question : First of all lets assume that $M_A$ is caused by force $F$ applied at distance $x$ from A, and distance from A to B is $L$.

To calculate $M_a$
$xF = M_A$
$M_a = \sqrt[2]{x^2 + L^2} F \sin\theta$
$M_a = \sqrt[2]{x^2 + L^2} F x / \sqrt[2]{x^2 + L^2}$
$M_a = M_A $
$M_a$ is moment of the force which provides moment $M_A$ about point A, about point B.

Rest of your equations look correct !