= Solution
The <symmetric matrix> of the <quadratic form> in the standard basis is
$$
A=\begin{pmatrix}
1&1&1\\
1&1&-1\\
1&-1&2
\end{pmatrix},
$$
because the off-diagonal entries contribute twice to $x^TAx$.
Apply <Gram-Schmidt orthogonalization for a symmetric bilinear form> to
$$
u_1=e_1,
\qquad
u_2=e_3-e_1,
\qquad
u_3=e_2-e_1+2u_2=-3e_1+e_2+2e_3.
$$
These vectors are pairwise orthogonal for $B(u,v)=u^TAv$, and
$$
B(u_1,u_1)=1,
\qquad
B(u_2,u_2)=1,
\qquad
B(u_3,u_3)=-4.
$$
Thus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrix
$$
\operatorname{diag}(1,1,-4).
$$
It follows that its <rank of a quadratic form>[rank] is $3$ and its <signature of a quadratic form>[signature] is
$$
\boxed{2-1=1}.
$$
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