The shortest distance between the lines

$${{x - 3} \over 3} = {{y - 8} \over { - 1}} = {{z - 3} \over 1}$$ and

$${{x + 3} \over { - 3}} = {{y + 7} \over 2} = {{z - 6} \over 4}$$ is :

If a, b and c are the greatest value of ¹⁹C_p, ²⁰C_q and ²¹C_r respectively, then :

Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :

If the equation, x² + bx + 45 = 0 (b $$ \in $$ R) has conjugate complex roots and they satisfy |z +1| = 2$$\sqrt {10} $$ , then :