If $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ are the position vectors of points $A, B, C$ and $D$ respectively,then find the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$. Deduce that $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are collinear.

(D) Let $\theta$ be the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$.
First,calculate the vectors:
$\overrightarrow{AB} = (2\hat{i} + 5\hat{j}) - (\hat{i} + \hat{j} + \hat{k}) = \hat{i} + 4\hat{j} - \hat{k}$
$\overrightarrow{CD} = (\hat{i} - 6\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} - 8\hat{j} + 2\hat{k}$
Calculate the magnitudes:
$|\overrightarrow{AB}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$
$|\overrightarrow{CD}| = \sqrt{(-2)^2 + (-8)^2 + 2^2} = \sqrt{4 + 64 + 4} = \sqrt{72} = 6\sqrt{2}$
Calculate the dot product:
$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(-2) + (4)(-8) + (-1)(2) = -2 - 32 - 2 = -36$
Calculate the angle $\cos \theta$:
$\cos \theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{-36}{(3\sqrt{2})(6\sqrt{2})} = \frac{-36}{18 \times 2} = \frac{-36}{36} = -1$
Since $\cos \theta = -1$,we have $\theta = \pi$.
Since the angle between the vectors is $\pi$,the vectors are collinear (specifically,they are in opposite directions). Alternatively,$\overrightarrow{CD} = -2(\hat{i} + 4\hat{j} - \hat{k}) = -2\overrightarrow{AB}$,which confirms they are collinear.