The shortest distance between the lines

$${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$$ and $${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$$, is :

Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area $$2\sqrt 2 $$. Let the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ be acute, $$|\overrightarrow a | = 1$$, and $$|\overrightarrow a \,.\,\overrightarrow b | = |\overrightarrow a \times \overrightarrow b |$$. If $$\overrightarrow c = 2\sqrt 2 \left( {\overrightarrow a \times \overrightarrow b } \right) - 2\overrightarrow b $$, then an angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ is :

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $${9 \over 4}$$ respectively. Then $${x^4} + {y^2}$$ is equal to :

If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x $$-$$ 6y = 30, then the probability that y < 1 is :