Hamming weight/distance; validate x = 0111000 with given H

Hamming weight

The Hamming weight $w(\mathbf{x})$ of a code vector is the number of 1s in it. For $\mathbf{x}=(0111000)$:

$$w(\mathbf{x})=0+1+1+1+0+0+0=3$$

Hamming distance

The Hamming distance between two code vectors is the number of positions in which they differ, $d(\mathbf{x},\mathbf{y})=w(\mathbf{x}\oplus\mathbf{y})$. For example, distance of $\mathbf{x}$ from the all-zero codeword is $d(0111000,0000000)=3$, and from $\mathbf{y}=1111111$ it is $w(1000111)=4$.

Proof that x is a valid codeword

A vector is a valid codeword of the code if and only if its syndrome is zero:

$$\mathbf{s}=H\mathbf{x}^T=\mathbf{0}\quad(\text{equivalently } \mathbf{x}H^T=\mathbf{0})$$ $$H=\begin{bmatrix}1&1&1&0&1&0&0\\1&1&0&1&0&1&1\\1&0&1&1&0&0&1\end{bmatrix},\qquad \mathbf{x}^T=\begin{bmatrix}0\\1\\1\\1\\0\\0\\0\end{bmatrix}$$

Only positions 2, 3 and 4 of $\mathbf{x}$ are 1, so $H\mathbf{x}^T$ is the modulo-2 sum of columns 2, 3 and 4 of $H$:

$$\mathbf{s}=H\mathbf{x}^T=\begin{bmatrix}0\\0\\0\end{bmatrix}$$

Since the syndrome is the zero vector, $\mathbf{x}=(0111000)$ satisfies all three parity-check equations. Hence the given code vector is valid.