Let C be a curve given by $y(x)=1+\sqrt{4 x-3}$, $x>\frac{3}{4}$. If P is a point on C , such that the tangent at P has slope $\frac{2}{3}$, then a point through which the normal at P passes, is

The equation of the circle, the end points of whose diameter are the centres of the circles $x^2+y^2+6 x-14 y+5=0$ and $x^2+y^2-4 x+10 y-4=0$ is

The equation of the plane, passing through the point $(1,1,1)$ and perpendicular to the planes $2 x+y-2 z=5$ and $3 x-6 y-2 z=7$, is

If for $x \in\left(0, \frac{1}{4}\right)$, the derivative of $\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)$ is $\sqrt{x} \cdot g(x)$, then $g(x)$ equals