Let a and b respectively be the semitransverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e² − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a² − b² is equal to :

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 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 :