The area (in sq. units) of the region bounded by the curve x² = 4y and the straight line x = 4y – 2 is :

A square is inscribed in the circle x² + y² – 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :

Let $${\left( { - 2 - {1 \over 3}i} \right)^3} = {{x + iy} \over {27}}\left( {i = \sqrt { - 1} } \right),\,\,$$ where x and y are real numbers, then y $$-$$ x equals :

Let A = $$\left( {\matrix{ 0 & {2q} & r \cr p & q & { - r} \cr p & { - q} & r \cr } } \right).$$   If  AA^T = I₃,   then   $$\left| p \right|$$ is :