Let the origin is shifted to $P(h, k)$ If new co-ordinates of any point $O\left(x^{\prime}, y^{\prime}\right)$
$$
\begin{aligned}
& x^{\prime}=x-h \Rightarrow x=x^{\prime}+h \\
& y^{\prime}=y-k \Rightarrow y=y^{\prime}+k \\
& \therefore \quad 3\left(x^{\prime}+h\right)^2+\left(y^{\prime}+k\right)^2-6\left(x^{\prime}+h\right)+4\left(y^{\prime}+k\right)+4=0 \\
& \Rightarrow 3\left(x^{\prime 2}+h^2+2 h x^{\prime}\right)+y^{\prime 2}+k^2+2 k y^{\prime}-6 x^{\prime} \\
& -6 h+4 y^{\prime}+4 k+4=0 \\
& \Rightarrow 3 x^{\prime 2}+y^{\prime 2}+(6 h-6) x^{\prime}+(2 k+4) y^{\prime} \\
& +\left(3 h^2+k^2-6 h+4 k+4\right)=0 \\
&
\end{aligned}
$$
$\Rightarrow$ To get equation with no 1 degree term,
$$
\begin{aligned}
& 6 h-6=0 \Rightarrow h=1 \\
& 2 k+4=0 \Rightarrow k=-2
\end{aligned}
$$

Hence required translation is,
$$
\begin{aligned}
& x=x^{\prime}+h=x^{\prime}+1 \\
& y=y^{\prime}+k=y^{\prime}+2
\end{aligned}
$$
putting this translation in given equation
$3 x^2+y^2-6 x+4 y+4=0$ we get
$$
2 x^2+3 x y-5 y^2=0
$$