If the points A, B, C with position vectors $20\hat i+λ\hat j, 5\hat i-\hat j$ and $10\hat i-13\hat j$ respectively are collinear, then the value of $λ$ is

-37

The correct answer is Option (2) → -37

Let

$A(20,\,\lambda),\; B(5,\,-1),\; C(10,\,-13)$

Vectors:

$\overrightarrow{AB}=(5-20,\,-1-\lambda)=(-15,\,-1-\lambda)$

$\overrightarrow{AC}=(10-20,\,-13-\lambda)=(-10,\,-13-\lambda)$

For collinearity,

$\displaystyle \frac{-15}{-10}=\frac{-1-\lambda}{-13-\lambda}$

$\displaystyle \frac{3}{2}=\frac{1+\lambda}{13+\lambda}$

Cross–multiply:

$3(13+\lambda)=2(1+\lambda)$

$39+3\lambda=2+2\lambda$

$\lambda=-37$

The value of $\lambda$ is $-37$.