Matlab, operator A\B What is the result of the operation A\B, where A(1, m) and B (1, m)?
In the manual it is written: 
A\B returns a least-squares solution to the system of equations A*x= B.

So it means x = inv (A'*A)*A'*B? However, the matrix A'*A is singular...
Let us suppose:
A=[1 2 3]
B=[6 7 6]

Then
A\B

0         0         0
0         0         0
2.0000    2.3333    2.0000

If ve use MLS:
C = inv (A'*A)   singular matrix
C = pinv(A'*A)

0.0051    0.0102    0.0153
0.0102    0.0204    0.0306
0.0153    0.0306    0.0459

D= C*A'*B

0.4286    0.5000    0.4286
0.8571    1.0000    0.8571
1.2857    1.5000    1.2857

So results A\B and inv (A'*A)*A'*B are different...
 A: If you have $A$ and $B$ as both row vectors, then C=A\B computes the matrix $C$ such that
$$C=A\backslash B \implies AC=B.$$
Recognize that since $A$ and $B$ are $1\times m$ row vectors, $C$ will be an $m\times m$ matrix. This matrix is not unique. One way to construct such a matrix is to zero out all but one of the rows. Say the $k$th row is non-zero. Populate the entry in the $i$th column such that $C_{ki}$ is equal to $b_i/a_k$. Then, your vector-matrix multiplication yields, for the $i$th entry of the resulting vector, the product
$$\sum_{n=1}^m a_n C_{ni} = a_kC_{ki} = b_i$$
because $C_{ni} = 0$ if $n \neq k$.
This result, as it turns out, is quite uninteresting.
A: In the interesting cases I knew before answering this question -the ones which appear in real problems I'm aware of-, the matrix $A$ is just the opposite of the one you've chosen: it usually has (way far) more rows than columns (see overdetermined system) and those columns are linearly independent vectors. Let's call them $a_1, \dots , a_n$:
$$
A = (a_1 \dots a_n) \ .
$$
Then $A^tA$ is the matrix of dot products
$$
A^tA = 
\begin{pmatrix}
a_1\cdot a_1  & \dots  & a_1\cdot a_n \\
\vdots        & \vdots & \vdots    \\
a_n\cdot a_1  & \dots  & a_n\cdot a_n
\end{pmatrix}
$$
which is always non-singular.
EDIT. In your case, your matrix $A$ has three non-linearly independent columns. All of your systems $x +2y + 3z = 6, 7, 6$ have infinitely many solutions. Matlab choses one of them.  For instance, for the first system $x + 2y + 3z = 6$, my Matlab gives me $(0,0,2)$ as a solution an explains to us 

help \ backslash
If A is an M-by-N matrix with M < or > N and B is a column
     vector with M components, or a matrix with several such columns,
     then X = A\B is the solution in the least squares sense to the
     under- or overdetermined system of equations A*X = B. The
     effective rank, K, of A is determined from the QR decomposition
     with pivoting. A solution X is computed which has at most K
     nonzero components per column. If K < N this will usually not
     be the same solution as PINV(A)*B.

So, indeed, Matlab has given a solution with at most $K=1 = \mathrm{rank}(A)$ nonzero components per column. Otherwise said, if I'm not wrong, Matlab has solved the linear system $x + 2y + 3z = 6, x= 0, y = 0$.
