There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

If A and B are two events such that $P(A) = 0.7$, $P(B) = 0.4$ and $P(A \cap \overline{B}) = 0.5$, where $\overline{B}$ denotes the complement of B, then $P\left(B \mid (A \cup \overline{B})\right)$ is equal to

Let $ \vec{a} = \hat{i} + 2\hat{j} + \hat{k} $ and $ \vec{b} = 2\hat{i} + \hat{j} - \hat{k} $. Let $ \hat{c} $ be a unit vector in the plane of the vectors $ \vec{a} $ and $ \vec{b} $ and be perpendicular to $ \vec{a} $. Then such a vector $ \hat{c} $ is:

Let the function $ f(x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0 $ be strictly increasing in $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $ \sum\limits_{i=1}^{5} \alpha_i^2 $ is equal to