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 :

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    $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to :

The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $$2\sqrt 2 $$ units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by :