The general solution of $\sin x-3 \sin 2 x+\sin 3 x=\cos x-3 \cos 2 x+\cos 3 x$ is

The length of the perpendicular from the point $\mathrm{A}(1,-2,-3)$ on the line $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}$ is

The straight line, $2 x-3 y+17=0$ is perpendicular to the line passing through the points $(7,17)$ and $(15, \beta)$, then $\beta$ equals

If $\mathrm{f}(x)=\frac{x^2-x}{x^2+2 x}$ then $\frac{\mathrm{d}}{\mathrm{d} x}\left(\mathrm{f}^{-1}(x)\right)$ at $x=2$ is