$\begin{aligned} & \text y=x^3-10 x^2+31 x-30 \\ & \frac{d y}{d x}=3 x^2-20 x+31 \\ & \therefore \quad \text { Slope of normal }=\frac{-1}{\left(\frac{d y}{d x}\right)}\end{aligned}$
$$
\begin{aligned}
& \therefore \frac{-1}{3 x^2-20 x+31}=\frac{-1}{14} \\
& \Rightarrow 3 x^2-20 x+31=14 \\
& \Rightarrow(3 x-17)(x-1)=0 \Rightarrow x=1, \frac{17}{3} \\
& \because \quad x \in I \Rightarrow x=1 \\
& \therefore y=(1)^3-10(1)^2+31(1)-30=1-10+31-30 \\
& \therefore y=-8
\end{aligned}
$$

Slope of tangent at $(1,-8)=3(1)^2-20(1)+31=14$
$\therefore$ Equation of tangent: $y+8=14(x-1)$
For $x$ intercept, $y=0$
$$
\therefore \frac{8}{14}=x-1 \Rightarrow x=\frac{11}{7} \text {. }
$$