The short answer is that you get the syndrome $H_m$ by multiplying the received vector $r$ with the parity check matrix: $H_m=H r^T$.
There are several equivalent parity check matrices for this Hamming code, and you haven't shown us which is the one your source uses. The bits that you did give hint at the possibility that the check matrix could be
$$
H=\pmatrix{1&0&1&0&1&0&1\cr0&1&1&0&0&1&1\cr0&0&0&1&1&1&1\cr}.
$$
To be fair, that is one of the more popular choices, and also fits your received word / syndrome pair :-). The reason I'm nitpicking about this is that any one of the $7!=5040$ permutations of the columns would work equally well for encoding/decoding. Being an algebraist at heart, I often prefer an ordering that exhibits the cyclic nature of the code better. Somebody else might insist on an ordering the matches with systematic encoding. Doesn't matter much!
Here your received vector $r=(1,0,0,1,0,0,1)$ has bits on at positions 1, 4 and 7, so the syndrome you get is
$$
H_m=H r^T=\pmatrix{1\cr 0\cr0\cr}+\pmatrix{0\cr 0\cr1\cr}+\pmatrix{1\cr 1\cr1\cr}=\pmatrix{0\cr 1\cr0\cr}
$$
the modulo two sum of the first, fourth and seventh columns of $H$.
If $r$ were a valid encoded message, then the syndrome $H_m$ would be the zero vector. As that was not the case here, an error (one or more) has occurred. The task of the decoder is to find the most likely error, and the usual assumptions lead us to the goal of toggling as few bits of $r$ as possible.
The nice thing about the Hamming code is that we can always do this by toggling at most one bit. We identify $H_m$ as the second column of $H$, so to make the syndrome zero by correcting a single bit we need to toggle the second bit of $r$.
What makes the Hamming code tick is that all possible non-zero syndrome vectors occur as columns of $H$. Therefore we always meet our goal of making the syndrom zero by toggling (at most) a single bit of any received word.
Let $c(x) = \sum_{i=0}^{N-1} c_i x^i$ denote the polynomial corresponding
to the codeword $\mathbf c = [c_0, c_1, \ldots, c_{N-1}]$ of a cyclic binary Hamming
code with generator polynomial $g(x)$. Note that $N = 2^m-1$ for some
positive integer $m$ and that $g(x)$ is a primitive
polynomial of degree $m$. Let $\alpha \in \mathbb F_{2^m}$ be a root of
$g(x)$ and note that $\alpha$ has order $2^m-1$.
The parity check matrix $H$ of this cyclic binary Hamming code can be viewed as
$[1, \alpha, \cdots, \alpha^{N-1}]$. If you are unfamiliar with this notion, then
express each $\alpha^i$ as a column vector to get the
usual binary representation. But, with this formulation,
$H\mathbf c^T = \mathbf0$ is equivalent to $c(\alpha) = 0$ which is
exactly what the polynomial formulation says in other terms: each
codeword polynomial is a multiple of $g(x)$ which happens to have
$\alpha$ as a root, and so $c(\alpha) = 0$.
Let $\mathbf c$ be transmitted and $\mathbf r = \mathbf c + \mathbf e$ denote
the received word. In terms of polynomials, $r(x) = \sum_i r_ix^i$
is the received word polynomial where $r_i = c_i + e_i$ with
$e(x)$ denoting the error word polynomial corresponding to the error
word $\mathbf e$. Note that $r(x) = c(x) + e(x)$. The receiver computes
$$H\mathbf r^T = H(\mathbf c + \mathbf e)^T =
H\mathbf c^T + H\mathbf e^T = \mathbf 0 + H\mathbf e^T = H\mathbf e^T.$$
In terms of polynomials, computing $H\mathbf r^T$ is the same as
evaluating $r(x)$ at $\alpha$, andso the above equation can be
expressed as
$$r(\alpha) = c(\alpha) + e(\alpha) = 0 + e(\alpha) = e(\alpha).$$
If a single error has occurred, then $\mathbf e$ has Hamming weight $1$,
and $e(x)$ is a monomial $x^i$ for some $i$, $0 \leq i \leq N-1$.
Thus, the bit in the $i$-th position has been received incorrectly.
In terms of polynomials, the syndrome is
$e(\alpha) = x^i\bigr |_{x=\alpha^i} = \alpha^i$. If two errors
have occurred, the syndrome is $\alpha^i + \alpha^{i^\prime} = \alpha^j$.
The decoder for the cyclic Hamming code cannot tell whether one
or two errors have occurred: if the symdrome equals $\alpha^\ell$,
it simply inverts $r_\ell$ to correct the error.
The extended Hamming code has an overall parity bit $c_N$
whose value is chosen to ensure that the extended codeword
$\hat{\mathbf c} = (\mathbf c, c_N)$ has even Hamming weight.
Suppose that an extended Hamming code is being used and
it is known that if there are two channel errors,
then they are adjacent errors not involving the overall
parity bit.
Let $\hat{\mathbf r} = (\mathbf r, r_N)$ be the received word for
the extended Hamming code. If $\hat{\mathbf r}$ has odd Hamming
weight, then the decoder assumes that a single error has
occurred, and if $r(\alpha) = \alpha^\ell$, the decoder inverts
$r_\ell$.
But if $\hat{\mathbf r}$ has even Hamming weight and
$r(\alpha) = \alpha^\ell \neq 0$, then the decoder assumes that
$e(x) = x^i + x^{i+1}$ (double adjacent error) and so
$r(\alpha) = e(\alpha) = \alpha^i + \alpha^{i+1} = \alpha^\ell.$
Since $\alpha^i = \alpha^\ell(1+\alpha)^{-1}$, the
decoder can find the value of $i$ and then invert $r_i$ and $r_{i+1}$.
In this way, single errors and double adjacent errors not involving
the overall parity bit can be corrected with an extended Hamming code.
Best Answer
That is correct. The off-diagonal bits are not protected at all in this scheme, and toggling one of them does not affect any other bits. Therefore the minimum distance is $1$.