= Solution
The <Gram-Schmidt process> first gives $u_1=a_1=(1,1,1,1)^T$, of norm two. Next $a_1^Ta_2/(a_1^Ta_1)=2$, so
$$
u_2=a_2-2u_1=(1,-1,-1,1)^T.
$$
For the third column, both coefficients onto $u_1,u_2$ equal one, giving
$$
u_3=a_3-u_1-u_2=(1,-1,1,-1)^T.
$$
These three nonzero vectors are mutually <orthogonal> and each has norm two. The resulting <orthonormal basis> is
$$
\boxed{q_1=\frac12\begin{pmatrix}1\\1\\1\\1\end{pmatrix},\quad
q_2=\frac12\begin{pmatrix}1\\-1\\-1\\1\end{pmatrix},\quad
q_3=\frac12\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix}.}
$$
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