= Solution
The two prescribed directions have <dot product> zero. Normalize them to
$$
\mathbf q_1=\frac{(1,1,1)^T}{\sqrt3},\qquad
\mathbf q_2=\frac{(0,-1,1)^T}{\sqrt2}.
$$
Their <cross product> is the perpendicular <unit vector> $\mathbf q_3=(2,-1,-1)^T/\sqrt6$. Using these as columns gives
$$
\boxed{Q=\begin{pmatrix}
1/\sqrt3&0&2/\sqrt6\\
1/\sqrt3&-1/\sqrt2&-1/\sqrt6\\
1/\sqrt3&1/\sqrt2&-1/\sqrt6
\end{pmatrix}}.
$$
The columns form an <orthonormal basis>, so $Q^TQ=I$ and $Q$ is an <orthogonal matrix>.
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