The point A, with coordinates (0, a, b) lies on the line l₁, which has equation r = 6i + 19j - k + λ(i + 4j - 2k) - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 6

Next, we need to find the position vector of point P. As given, OP is perpendicular to l₁. The direction vector of the line l₁ is obtained from the given line equation:

Direction vector, d = (1, 4, -2).

For OP to be perpendicular to l₁, their dot product must equal zero:

Let OP = (x, y, z), then:

$ext{If } OP = (6 - λ, 19 + 4λ, -1 - 2λ), ext{ we can substitute } λ = -6.$

This results in:

$ext{Thus, OP = (6 - (-6), 19 + 4(-6), -1 - 2(-6)) = (12, -5, 11)}.$

Therefore, the position vector of point P is:

$P = 2i + 3j + 7k.$

To show that points A, P, and B are collinear, we need to check if the vectors AP and PB are scalar multiples of each other.

Vector AP can be found as:

$AP = P - A = (2i + 3j + 7k) - (0i - 5j + 11k) = 2i + 8j - 4k.$

Vector PB can be calculated from:

$B - P = (5i + 15j + 1k) - (2i + 3j + 7k) = (5 - 2)i + (15 - 3)j + (1 - 7)k = 3i + 12j - 6k.$

Next, to check for collinearity, we check if there exists a scalar k such that: