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 $x + 2y + 3z = 6, x= 0, y = 0$.