The distance of the point $$\mathrm{P}(-2,4,-5)$$ from the line $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$ is

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is

$$\int \frac{x^2+1}{x\left(x^2-1\right)} \mathrm{d} x=$$

If the matrix $$\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -5 & 1\end{array}\right]$$ and $$\mathrm{A}^{-1}=x \mathrm{~A}+y \mathrm{I}$$, when $$I$$ is a unit matrix of order 2 , then the value of $$2 x+3 y$$ is