We need to show that the four points $A = -6i + 3j + 2k$, $B = 3i - 2j + 4k$, $C = 5i + 7j + 3k$, and $D = -13i + 17j - k$ are coplanar.

1. Problem Description

We need to show that the four points $A = -6i + 3j + 2k$, $B = 3i - 2j + 4k$, $C = 5i + 7j + 3k$, and $D = -13i + 17j - k$ are coplanar.

2. Solution Steps

To show that four points are coplanar, we can show that the three vectors formed by these points are coplanar. Let's find the vectors $AB$, $AC$, and $AD$.

$AB = B - A = (3i - 2j + 4k) - (-6i + 3j + 2k) = (3 - (-6))i + (-2 - 3)j + (4 - 2)k = 9i - 5j + 2k$

$AC = C - A = (5i + 7j + 3k) - (-6i + 3j + 2k) = (5 - (-6))i + (7 - 3)j + (3 - 2)k = 11i + 4j + k$

$AD = D - A = (-13i + 17j - k) - (-6i + 3j + 2k) = (-13 - (-6))i + (17 - 3)j + (-1 - 2)k = -7i + 14j - 3k$

The points $A$, $B$, $C$, and $D$ are coplanar if the scalar triple product of the vectors $AB$, $AC$, and $AD$ is equal to zero. The scalar triple product is given by the determinant of the matrix formed by the components of the vectors.

Scalar triple product = $AB \cdot (AC \times AD) = \begin{vmatrix} 9 & -5 & 2 \\ 11 & 4 & 1 \\ -7 & 14 & -3 \end{vmatrix}$

$= 9 \begin{vmatrix} 4 & 1 \\ 14 & -3 \end{vmatrix} - (-5) \begin{vmatrix} 11 & 1 \\ -7 & -3 \end{vmatrix} + 2 \begin{vmatrix} 11 & 4 \\ -7 & 14 \end{vmatrix}$

$= 9(4(-3) - 1(14)) + 5(11(-3) - 1(-7)) + 2(11(14) - 4(-7))$

$= 9(-12 - 14) + 5(-33 + 7) + 2(154 + 28)$

$= 9(-26) + 5(-26) + 2(182)$

$= -234 - 130 + 364$

$= -364 + 364$

$= 0$

Since the scalar triple product is zero, the vectors $AB$, $AC$, and $AD$ are coplanar. Therefore, the points $A$, $B$, $C$, and $D$ are coplanar.

3. Final Answer

The four points are coplanar.