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

The tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8 x+6 y+20=0$ then the co-ordinates of the corresponding point of contact is

The general solution of the differential equation $x \cos y \mathrm{~d} y=\left(x \mathrm{e}^{\mathrm{x}} \log x+\mathrm{e}^x\right) \mathrm{d} x$ is given by

The equation $(\operatorname{cosp}-1) x^2+(\cos p) x+\operatorname{sinp}=0$ in the variable $x$, has real roots. Then p can take any value in the interval