Let the tangent and normal at the point $$(3 \sqrt{3}, 1)$$ on the ellipse $$\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$$ meet the $$y$$-axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$A B$$ as a diameter and the line $$x=2 \sqrt{5}$$ intersect $$C$$ at the points $$P$$ and $$Q$$. If the tangents at the points $$P$$ and $$Q$$ on the circle intersect at the point $$(\alpha, \beta)$$, then $$\alpha^{2}-\beta^{2}$$ is equal to :

The distance of the point $$(-1,2,3)$$ from the plane $$\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$$ and $$\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$$ is :

For $$x \in \mathbb{R}$$, two real valued functions $$f(x)$$ and $$g(x)$$ are such that, $$g(x)=\sqrt{x}+1$$ and $$f \circ g(x)=x+3-\sqrt{x}$$. Then $$f(0)$$ is equal to

For the differentiable function $$f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$$, let $$3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$$, then $$\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$$ is equal to