SOLUTION The vector \(\vec{PQ} \times \vec{PR}\) is perpendicular
to both \(\vec{PQ}\) and \(\vec{PR}\) and is therefore perpendicular
to the plane through P, Q, and R. We know from (12.2.1) that \[
\vec{PQ} = (-2 - 1)\mathbf{i} + (5 - 4)\mathbf{j} + (-1 - 6)\mathbf{k} =
-3\mathbf{i} + \mathbf{j} - 7\mathbf{k}
\] \[
\vec{PR} = (1 - 1)\mathbf{i} + (-1 - 4)\mathbf{j} + (1 - 6)\mathbf{k} =
-5\mathbf{j} - 5\mathbf{k}
\] We compute the cross product of these vectors: \[
\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j}
& \mathbf{k} \\ -3 & 1 & -7 \\ 0 & -5 & -5
\end{vmatrix}
\] \[
= (-5 - 35)\mathbf{i} - (15 - 0)\mathbf{j} + (15 - 0)\mathbf{k} =
-40\mathbf{i} - 15\mathbf{j} + 15\mathbf{k}
\]
So the vector \(\langle -40, -15, 15
\rangle\) is perpendicular to the given plane. Any nonzero scalar
multiple of this vector, such as \(\langle -8,
-3, 3 \rangle\), is also perpendicular to the plane.
