Question: Consider the following system of three equations:
$$2y+2z=9-2x$$ $$x=12-3y-4z$$ $$Ax+5y+6z=B$$
Find values of A and B which makes the system of equations inconsistent and then give a geometric interpretation (sketch a diagram)
What I have attempted:
$$2y+2z=9-2x ⇔ 2x+2y+2z=9$$ $$x=12-3y-4z ⇔ x+3y+4z=12$$ $$Ax+5y+6z=B$$
Using the first 2 equations to find y in terms of x $$2x+2y+2z=9$$
$$x+3y+4z=12$$
To eliminate z multiply first equation by 2
$$4x+4y+4z=18$$
$$x+3y+4z=12$$
$$ 3x+y = 6 ⇔ \mathbf{y=6-3x}$$
To eliminate y multiply first equation by 3 and multiply second equation by 2
$$6x+6y+6z=27$$
$$2x+6y+8z=24$$
$$ 4x-2z=3 ⇔ \mathbf{z=2x-\frac{3}{2}}$$
Subbing in $\mathbf{y=6-3x}$ and $\mathbf{z=2x-\frac{3}{2}}$ into $$Ax+5y+6z=B$$
$$ Ax+5(6-3x) + 6(2x-\frac{3}{2}) = B $$
$$ Ax-3x+21 = B $$
$$ (A-3)x + 21 = B $$
To have a system inconsistent (x=0)
$$ (A-3) = 0 ⇔ A = 3 $$
$$ A = 3 $$ $$ B = 21 $$
(Please check If i am correct) and if so how would I make a geometric representation of this inconsistent system of equations?
I searched google and got this 
but I'm not too sure if it is correct.. thanks for the help!
A key point is to rewrite your system into a matrix form, then examine the determinant $D$ of your system. it is $D=2A-6$.
If $D$ is nonzero, there is a unique solution (your planes intersect into a single point).
If $D=0$, i.e. for $A=3$, look at your system. Assume that it has a solution $(x,y,z)$. Visibly, if you add the first 2 equations and substract the last one, you obtain $0=21-B$.
Either $B \neq 21$, no solution can exist, by contradiction.
Or $B=21$, in which case the third equation, being a consequence of the first 2 can be cancelled. You are left with two planes which intersect into a line representing an infinite number of solutions (this case, three planes intersecting along a common line, is not represented in your graphics)
Edit: As you are still in high school, in order to understand what is a determinant (a central concept/tool in mathematics) and how it can be used, I advise you to begin by http://www.mathnstuff.com/math/algebra/adeterm.htm which intrduces the two dimensional case. Then proceed to the article ("Applications" part) https://en.wikipedia.org/wiki/Cramer's_rule treating as well the completely similar 3 dimensional case, knowing that a $3 \times 3$ determinant can be computed by Sarrus' method https://en.wikipedia.org/wiki/Rule_of_Sarrus. This method is effective and gives a UNIQUE solution if the determinant of the system is non zero (it is the common denominator). Said otherwise, exceptional cases can only happen when the system's determinant is zero.