= Solution
For a real square <matrix>, a <QR factorization> writes $B=QR$ with $Q$ an <orthogonal matrix>, $Q^TQ=I$, and $R$ an <upper triangular matrix>. If $B$ is invertible, so is $R$, and
$$
Bx=b\quad\Longleftrightarrow\quad Rx=Q^Tb.
$$
Compute $Q^Tb$ and use <back substitution>; no <matrix inverse> needs to be formed.
Apply the <Gram-Schmidt process> to the columns $b_1,b_2,b_3$. The first column has <Euclidean norm> two, giving $q_1=(1,0,0)^T$ and $r_{11}=2$. For the second column,
$$
r_{12}=q_1^Tb_2=1,\qquad b_2-r_{12}q_1=(0,4,3)^T,
\qquad r_{22}=5,\qquad q_2=(0,4/5,3/5)^T.
$$
For the third column,
$$
r_{13}=q_1^Tb_3=0,\qquad r_{23}=q_2^Tb_3=7,
\qquad b_3-7q_2=(0,-3/5,4/5)^T.
$$
This residual already has <Euclidean norm> one. Set $q_3=(0,-3/5,4/5)^T$ and $r_{33}=1$. Thus
$$
\boxed{Q=\begin{pmatrix}1&0&0\\0&4/5&-3/5\\0&3/5&4/5\end{pmatrix},\qquad
R=\begin{pmatrix}2&1&0\\0&5&7\\0&0&1\end{pmatrix}.}
$$
The columns of $Q$ are <orthonormal>, and <matrix multiplication> verifies $QR=B$.
Using precisely this <QR factorization>, transform the right-hand side:
$$
Q^T\begin{pmatrix}-3\\1\\2\end{pmatrix}
=\begin{pmatrix}-3\\(4+6)/5\\(-3+8)/5\end{pmatrix}
=\begin{pmatrix}-3\\2\\1\end{pmatrix}.
$$
<Back substitution> in $Rx=Q^Tb$ gives
$$
x_3=1,\qquad 5x_2+7=2\ \Longrightarrow\ x_2=-1,
\qquad 2x_1-1=-3\ \Longrightarrow\ x_1=-1.
$$
Therefore
$$
\boxed{x=(-1,-1,1)^T.}
$$
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