I'm trying to understand how (X2 -6x + _) fills as (X2 -6x + 9), also, how (2 + 9 + 1/4 + 9/4) is somehow (X-3)2+ (Y +1/2)2+ (Z - 3/2) and =...

This process is called solve by completing the square. All
you need to to is rewrite the equation so you will create a complete square that you can
write as one term.

For
example:

We have the
equation:

x^2 + 2x + 7 = 0

We
need to rewrite as a completer square.

You will use (x^2 +
2x) and complete the square.

Then you will need to add the
coefficient of x /2 which is2/2 = 1

You will also need to
add 1 to the left side so the equality remains the
same.

Then we will add 1
:

==> x^2 + 2x + 1 + 7 = 
1

==> Now we can write the first three terms as a
completer square.

==> (x+1)^2 + 7 =
1

Now we will subtract 1 from both
sides.

==> (x+1)^2 + 6 =
0

Then we conclude that:

x^2 +
2x +7 = (x+1)^2 + 6

Now we will try and solve the examples
yoy provided.

x^2 + y^2 + z^2 -6x + y -3z -2 =
0

First we will group terms with the same
letter.

==> (x^2 -6x) + (y^2 +y) + (z^2 -3z) -2=
0

Now we will complete the square for each
term

For (x^2 -6x) we will add (6/2)^2 = 3^2 = 9  to both
sides.

For (y^2 +y) we will add (1/2)^2 = 1/4  to both
sides

For (z^2 -3z) we will add (3/2)^2 = 9/4 to both
sides.

==> (x^2 - 6x +9) + (y^2+y +1/4) + (z^2 - 3z
+ 9/4) -2 = 9 + 1/4 + 9/4

==> Now we will rewrite as
a complete squares.

==> (x-3)^2 + (y+1/2)^2 +
(z-3/2)^2  -2 = 9+ 10/4

==> (x-3)^2 + (y+1/2)62 +
(x-3/2)^2  = 9 + 10/4 + 2

==> (x-3)^2
+ (y+1/2)^2 + (x -3/2)^2 =  54/4 = 27/2