The points (3, −2, −1), (−1, 1, 2), (2, 3, −4) and (4, 5, λ) are coplanar when λ=

1. Break Down the Problem

To determine the value of $\lambda$ that makes the points coplanar, we can use the property that the volume of the parallelepiped formed by the vectors to these points should be zero. This can be accomplished by evaluating the determinant of a matrix formed by these points.

2. Relevant Concepts

For points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$, and $D(x_4, y_4, z_4)$, the points are coplanar if the determinant of the matrix formed by their coordinates (with an additional row of ones) equals zero.

The matrix is defined as:

$\text{Det} = \begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix}$

For our specific points, we have:

$\begin{vmatrix} 3 & -2 & -1 & 1 \\ -1 & 1 & 2 & 1 \\ 2 & 3 & -4 & 1 \\ 4 & 5 & \lambda & 1 \end{vmatrix} = 0$

3. Analysis and Detail

We will compute the determinant step by step.

$\text{Det} = \begin{vmatrix} 3 & -2 & -1 & 1 \\ -1 & 1 & 2 & 1 \\ 2 & 3 & -4 & 1 \\ 4 & 5 & \lambda & 1 \end{vmatrix}$

Using the cofactor expansion along the last row:

$= 4 \cdot \begin{vmatrix} -2 & -1 & 1 \\ 1 & 2 & 1 \\ 3 & -4 & 1 \end{vmatrix} - 5 \cdot \begin{vmatrix} 3 & -1 & 1 \\ -1 & 2 & 1 \\ 2 & -4 & 1 \end{vmatrix} + \lambda \cdot \begin{vmatrix} 3 & -2 & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -2 & -1 \\ -1 & 1 & 2 \\ 2 & 3 & -4 \end{vmatrix}$

Calculating each of the smaller determinants leads us to the following equations. For brevity, let's compute these values one by one:

1. Calculate the first determinant for the factor $4$:

$\begin{vmatrix} -2 & -1 & 1 \\ 1 & 2 & 1 \\ 3 & -4 & 1 \end{vmatrix} = -2(2 - (-4)) - (-1)(1 - 3) + 1(-8 + 3) = -2(6) + 2 - 5 = -12 + 2 - 5 = -15$

2. Calculate the second determinant for the factor $-5$:

$\begin{vmatrix} 3 & -1 & 1 \\ -1 & 2 & 1 \\ 2 & -4 & 1 \end{vmatrix} = 3(2 - 4) - (-1)(-1 - 4) + 1(4 + 2) = 3(-2) - 1(-5) + 1(6) = -6 + 5 + 6 = 5$

3. Calculate the third determinant for the factor $\lambda$:

$\begin{vmatrix} 3 & -2 & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} = 3(1 - 3) - (-2)(-1 - 2) + 1(-3 + 2) = 3(-2) - 2(-3) + 1(-1) = -6 + 6 - 1 = -1$

4. Finally, calculate the fourth determinant for the factor $-1$:

$\begin{vmatrix} 3 & -2 & -1 \\ -1 & 1 & 2 \\ 2 & 3 & -4 \end{vmatrix} = 3(1 \cdot -4 - 2 \cdot 3) - (-2)(-1 \cdot -4 - 2 \cdot 2) + (-1)(-1 \cdot 3 - 2 \cdot 2)$

Calculating:

$= 3(-4 - 6) + 2(-4 - 4) - 1(-3 - 4) = 3(-10) - 16 + 7 = -30 - 16 + 7 = -39$

4. Verify and Summarize

Putting it all together:

$4(-15) - 5(5) + \lambda(-1) - (-39) = 0 \\ -60 - 25 - \lambda + 39 = 0 \\ -46 - \lambda = 0 \\ \lambda = -46$

Final Answer

The value of $\lambda$ for the points to be coplanar is $\lambda = -46$.