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

The correct option is -37.

To find the value of $\lambda$ such that the points A, B, and C are collinear, we represent their position vectors as:
$\overrightarrow{a}=20\hat{i}+\lambda \hat{j}$
$\overrightarrow{b}=5\hat{i}-\hat{j}$
$\overrightarrow{c}=10\hat{i}-13\hat{j}$

Three points A, B, and C are collinear if the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel (collinear). This means one vector is a scalar multiple of the other:
$\overrightarrow{AB}=k\overrightarrow{BC}$
for some scalar $k$.

Let us first find the components of the vector $\overrightarrow{AB}$:
$\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}=(5-20)\hat{i}+(-1-\lambda )\hat{j}=-15\hat{i}-(1+\lambda )\hat{j}$

Next, let us find the components of the vector $\overrightarrow{BC}$:
$\overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}=(10-5)\hat{i}+(-13-(-1\left)\right)\hat{j}=5\hat{i}-12\hat{j}$

Since $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear, their corresponding components must be proportional:
$\frac{-15}{5}=\frac{-(1+\lambda )}{-12}$

Simplifying both sides of the equation:
$-3=\frac{1+\lambda }{12}$

Multiply both sides by 12:
$-36=1+\lambda$

Subtracting 1 from both sides gives the value of $\lambda$:
$\lambda =-36-1=-37$

Thus, the value of $\lambda$ is -37.