Let the required point be $P\left(x_{1}, y_{1}\right)$.
The equation of the given curve is $y=3 x^{2}-11 x-15$
$\Rightarrow \frac{d y}{d x}=6 x-11$
$\Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=6 x_{1}-11$
Since, the tangent at $P$ is parallel to the line joining $(5,5)$ and $(11,227)$
$\therefore$ Slope of the tangent at $P=$ Slope of the line joining $(5,5)$ and $(11,227)$ $\Rightarrow \left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{227-5}{11-5}$
$\Rightarrow 6 \,x_{1}-11=\frac{222}{6}$
$\Rightarrow 6 \,x_{1}-11=37$
$\Rightarrow 6 \,x_{1}=48 \Rightarrow x_{1}=8$