In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$,   $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies on a line :

If A and B are any two events such that P(A) = $${2 \over 5}$$ and P (A $$ \cap $$ B) = $${3 \over {20}}$$, hen the conditional probability, P(A $$\left| {} \right.$$(A' $$ \cup $$ B')), where A' denotes the complement of A, is equal to :

If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is :

The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and
$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :