The equation of the normal to the curve \(3 x^2-y^2=8\), which is parallel to the line \(x+3 y=10\), is

\(
3 x^2-y^2=8
\)
Differentiating w.r.t. \(x\), we get
\(6 x-2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x}{y}\)
\(\therefore \) Slope of the tangent to the curve is \(\frac{3 x}{y}\).
\(\therefore \) Slope of the normal is \(\frac{-y}{3 x}\).
It is parallel to line \(x+3 y=10 \Rightarrow\) slope \(=-\frac{1}{3}\)
\(\therefore \frac{-y}{3 x}=\frac{-1}{3} \Rightarrow x=y\)
\(\therefore \) When \(x=y\), equation of the curve becomes
\(\therefore 3 x^2-x^2=8\)
\(\therefore x^2=4\)
\(\therefore x=2,-2 \Rightarrow y=2,-2\)
\(\therefore (2,2)\) and \((-2,-2)\) are the points of contact of the normal and the curve.
\(\therefore \) Equations are \((y-2)=\frac{-1}{3}(x-2)\) or
\(
\begin{aligned}
& (y+2)=\frac{-1}{3}(x+2) \\
& \text { i.e., } x+3 y-8=0 \text { or } x+3 y+8=0
\end{aligned}
\)