To determine the value of \(\lambda\) for which the points A, B, and C with the given position vectors are collinear, we need to check the condition for collinearity. The points are collinear if the vector AC is a scalar multiple of the vector AB.

The position vectors for the points are:

• Point A: Position Vector: \(20\hat{i} + \lambda\hat{j}\)
• Point B: Position Vector: \(5\hat{i} - \hat{j}\)
• Point C: Position Vector: \(10\hat{i} - 13\hat{j}\)

First, calculate the vectors AB and AC:

• Vector AB: \(\overrightarrow{AB} = \mathbf{B} - \mathbf{A} = \left(5\hat{i} - \hat{j}\right) - \left(20\hat{i} + \lambda\hat{j}\right)\)
• This simplifies to: \(\overrightarrow{AB} = -15\hat{i} - \left(1 + \lambda\right)\hat{j}\)
• Vector AC: \(\overrightarrow{AC} = \mathbf{C} - \mathbf{A} = \left(10\hat{i} - 13\hat{j}\right) - \left(20\hat{i} + \lambda\hat{j}\right)\)
• This simplifies to: \(\overrightarrow{AC} = -10\hat{i} - \left(13 + \lambda\right)\hat{j}\)

For the points to be collinear, Vector AC must be a scalar multiple of Vector AB. Thus, we equate:

\(k(-15\hat{i} - (1 + \lambda)\hat{j}) = -10\hat{i} - (13 + \lambda)\hat{j}\)

Comparing the \(\hat{i}\) and \(\hat{j}\) components, we get:

• From \(\hat{i}\): \(-15k = -10\), solves to \(k = \frac{10}{15} = \frac{2}{3}\)
• From \(\hat{j}\): \(-(1 + \lambda)k = -(13 + \lambda)\)

Substitute \(k = \frac{2}{3}\) in the \(\hat{j}\) equation:

\(-\left(1 + \lambda\right)\left(\frac{2}{3}\right) = -(13 + \lambda)\)

This simplifies to:

\(\frac{-2 - 2\lambda}{3} = -(13 + \lambda)\)

Cross-multiply and simplify:

\(-2 - 2\lambda = -39 - 3\lambda\)

Rearranging gives:

\(3\lambda - 2\lambda = -39 + 2\)

\(\lambda = -37 + 2 = 37\)

Therefore, the value of \(\lambda\) for which the points are collinear is 37.