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

Continuity requires $\lim_{x \to 3} g(x) = g(3)$, so $K = \lim_{x \to 3} \frac{\int_3^{f(x)} 3t^2\,dt}{x-3}$.

The limit is indeterminate ($0/0$), so apply L'Hôpital's Rule. Differentiate numerator and denominator separately.

Denominator derivative: $\frac{d}{dx}(x-3) = 1$.

Numerator derivative (via Leibniz rule): $\frac{d}{dx} \int_3^{f(x)} 3t^2\,dt = 3(f(x))^2 f'(x)$.

$K = \lim_{x \to 3} \frac{3(f(x))^2 f'(x)}{1} = 3(f(3))^2 f'(3)$.

Given $f(3)=3$ and $f'(3)=\frac{1}{27}$, so $K = 3 \cdot 9 \cdot \frac{1}{27} = 1$.

$\boxed{1}$