If the solution curve of the differential equation

$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $$3x + y - 29 = 0$$, is $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$, then $$a + b + c + d + k$$ is equal to :

The set of values of k, for which the circle $$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$ lies inside the fourth quadrant and the point $$\left( {1, - {1 \over 3}} \right)$$ lies on or inside the circle C, is :

Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :