Are the four points \(A(1, 2, 1)\), \(B(2, -3, 4)\), \(C(3, 4, -5)\), \(D(2, 3, -2)\) coplanar? Justify your answer.

Answer: Yes, the four points are coplanar.

\(\overline{AB} = \hat{i} - 5\hat{j} + 3\hat{k}\), \(\overline{AC} = 2\hat{i} + 2\hat{j} - 6\hat{k}\), \(\overline{AD} = \hat{i} + \hat{j} - 3\hat{k}\).

The points are coplanar if \([\overline{AB}\ \overline{AC}\ \overline{AD}] = 0\).

\([\overline{AB}\ \overline{AC}\ \overline{AD}] = \begin{vmatrix} 1 & -5 & 3 \\ 2 & 2 & -6 \\ 1 & 1 & -3 \end{vmatrix} = 1(-6 + 6) + 5(-6 + 6) + 3(2 - 2) = 0\).

(Indeed \(\overline{AC} = 2\,\overline{AD}\), so two rows are proportional.) Since the scalar triple product is zero, the vectors are coplanar and hence A, B, C, D are coplanar.