Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :
Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then $$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :
For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$ , where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :
Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :