For which $\lambda$ are $A(1, 2, 3)$, $B(2, 3, 5)$ and $C(\lambda, 5, 9)$ collinear?

Explanation

1. 01

   $\overrightarrow{AB} = \hat{i} + \hat{j} + 2\hat{k}$ and $\overrightarrow{AC} = (\lambda - 1)\hat{i} + 3\hat{j} + 6\hat{k}$ must be parallel.

2. 02

   The $\hat{j}$ and $\hat{k}$ parts show $\overrightarrow{AC} = 3\overrightarrow{AB}$, so $\lambda - 1 = 3$: $\lambda = 4$.