Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable function having $\mathrm{f}(3)=3, \mathrm{f}^{\prime}(3)=\frac{1}{27}$ and $\mathrm{g}(x)= \begin{cases}\int_3^{\mathrm{f}(x)} \frac{3 \mathrm{t}^2}{x-3} \mathrm{dt}, & \text { if } x \neq 3 \\ \mathrm{~K}, & \text { if } x=3\end{cases}$ is continuous at $x=3$, then $\mathrm{K}=$

The area bounded by the curve $y=4 x-x^2$ and X - axis in square units, is $\qquad$

The Cartesian equation of the plane $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ is

If the line $\frac{x-3}{2}=\frac{y+5}{-1}=\frac{z+2}{2}$ lies in the plane $\alpha x+3 y-z+\beta=0$, then values of $\alpha$ and $\beta$ respectively are ….