To show that the points \( A(4, -1, 2) \), \( B(-3, 5, 1) \), \( C(2, 3, 4) \), and \( D(1, 6, 6) \) are coplanar, we can use the determinant method. The points are coplanar if the volume of the tetrahedron formed by these points is zero, which can be determined using the following determinant:
\[
\begin{vmatrix}
x_1 - x_4 & y_1 - y_4 & z_1 - z_4 \\
x_2 - x_4 & y_2 - y_4 & z_2 - z_4 \\
x_3 - x_4 & y_3 - y_4 & z_3 - z_4
\end{vmatrix} = 0
\]
### Step 1: Define the points
Let:
- \( A(4, -1, 2) \) be \( (x_1, y_1, z_1) \)
- \( B(-3, 5, 1) \) be \( (x_2, y_2, z_2) \)
- \( C(2, 3, 4) \) be \( (x_3, y_3, z_3) \)
- \( D(1, 6, 6) \) be \( (x_4, y_4, z_4) \)
### Step 2: Calculate the differences
Now, we compute the differences from point \( D(1, 6, 6) \):
- \( x_1 - x_4 = 4 - 1 = 3 \)
- \( y_1 - y_4 = -1 - 6 = -7 \)
- \( z_1 - z_4 = 2 - 6 = -4 \)
- \( x_2 - x_4 = -3 - 1 = -4 \)
- \( y_2 - y_4 = 5 - 6 = -1 \)
- \( z_2 - z_4 = 1 - 6 = -5 \)
- \( x_3 - x_4 = 2 - 1 = 1 \)
- \( y_3 - y_4 = 3 - 6 = -3 \)
- \( z_3 - z_4 = 4 - 6 = -2 \)
### Step 3: Set up the determinant
The determinant becomes:
\[
\begin{vmatrix}
3 & -7 & -4 \\
-4 & -1 & -5 \\
1 & -3 & -2
\end{vmatrix}
\]
### Step 4: Calculate the determinant
Now we calculate the determinant:
\[
= 3 \begin{vmatrix}
-1 & -5 \\
-3 & -2
\end{vmatrix} - (-7) \begin{vmatrix}
-4 & -5 \\
1 & -2
\end{vmatrix} - 4 \begin{vmatrix}
-4 & -1 \\
1 & -3
\end{vmatrix}
\]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix}
-1 & -5 \\
-3 & -2
\end{vmatrix} = (-1)(-2) - (-5)(-3) = 2 - 15 = -13 \)
2. \( \begin{vmatrix}
-4 & -5 \\
1 & -2
\end{vmatrix} = (-4)(-2) - (-5)(1) = 8 + 5 = 13 \)
3. \( \begin{vmatrix}
-4 & -1 \\
1 & -3
\end{vmatrix} = (-4)(-3) - (-1)(1) = 12 + 1 = 13 \)
Substituting back into the determinant:
\[
= 3(-13) + 7(13) - 4(13) = -39 + 91 - 52 = 0
\]
### Conclusion
Since the determinant equals zero, the points \( A, B, C, \) and \( D \) are coplanar.
### Step 5: Find the equation of the plane
To find the equation of the plane passing through points \( A, B, \) and \( C \), we can use the determinant form:
\[
\begin{vmatrix}
x - 4 & y + 1 & z - 2 \\
-7 & 6 & -1 \\
-2 & 4 & 2
\end{vmatrix} = 0
\]
Calculating the determinant:
\[
= (x - 4) \begin{vmatrix}
6 & -1 \\
4 & 2
\end{vmatrix} - (y + 1) \begin{vmatrix}
-7 & -1 \\
-2 & 2
\end{vmatrix} + (z - 2) \begin{vmatrix}
-7 & 6 \\
-2 & 4
\end{vmatrix}
\]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix}
6 & -1 \\
4 & 2
\end{vmatrix} = (6)(2) - (-1)(4) = 12 + 4 = 16 \)
2. \( \begin{vmatrix}
-7 & -1 \\
-2 & 2
\end{vmatrix} = (-7)(2) - (-1)(-2) = -14 - 2 = -16 \)
3. \( \begin{vmatrix}
-7 & 6 \\
-2 & 4
\end{vmatrix} = (-7)(4) - (6)(-2) = -28 + 12 = -16 \)
Substituting back into the equation:
\[
= (x - 4)(16) - (y + 1)(-16) + (z - 2)(-16) = 0
\]
Expanding this gives:
\[
16x - 64 + 16y + 16 - 16z + 32 = 0
\]
Combining like terms:
\[
16x + 16y - 16z - 16 = 0
\]
Dividing through by 16:
\[
x + y - z - 1 = 0
\]
Thus, the equation of the plane is:
\[
x + y - z = 1
\]
