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

The four points (3, -2, -1), (-1, 1, 2), (2, 3, -4) and (4, 5, λ) are coplanar if the determinant of the matrix formed by their coordinates is zero.

The matrix is:

| 3 -2 -1 1 | | -1 1 2 1 | | 2 3 -4 1 | | 4 5 λ 1 |

We can calculate the determinant of this matrix using the method of minors and cofactors.

First, we expand along the first row:

= 3 * det(A) - (-2) * det(B) - 1 * det(C) + 1 * det(D)

where A, B, C, and D are 3x3 matrices obtained by removing the row and column of each element in the first row.

A = | 1 2 1 | B = | -1 2 1 | C = | -1 1 1 | D = | -1 1 2 | | 3 -4 1 | | 2 -4 1 | | 2 2 1 | | 2 3 1 | | 5 λ 1 | | 4 λ 1 | | 4 5 1 | | 4 5 λ |

Calculating the determinants of these matrices:

det(A) = 1*(-41 - λ1) - 2*(31 - 51) + 1*(3λ - 52) = -4 - λ - 22 + 25 + 3λ - 10 = 2λ - 8

det(B) = -1*(-41 - λ1) - 2*(21 - 51) + 1*(2λ - 54) = 4 + λ - 22 + 25 + 2λ - 20 = 3λ - 8

det(C) = -1*(-41 - 11) - 1*(21 - 51) + 1*(25 - 41) = 4 + 1 - 22 + 25 - 4 = 6

det(D) = -1*(3λ - 21) - 1*(2λ - 54) + 2*(25 - 41) = -3λ + 2 - 2λ + 20 + 2*10 - 8 = -5λ + 24

Substituting these back into the original equation:

= 3*(2λ - 8) - (-2)(3λ - 8) - 16 + 1*(-5λ + 24) = 6λ - 24 + 6λ - 16 - 6 - 5λ + 24 = 7λ - 22

Setting this equal to zero gives the condition for the points to be coplanar:

7λ - 22 = 0 7λ = 22 λ = 22/7

So, the points are coplanar when λ = 22/7.