Let $\overrightarrow{n}$ be the normal vector to the plane passing through $(4, -1, 2)$ and parallel to the given lines, then we know that $\overrightarrow{n}$ is perpendicular to both the lines.

Also, we know that a vector perpendicular to two vectors lie along the cross product of the two vectors.

Hence, $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 2& 3\end{array}\right|$

$\Rightarrow \overrightarrow{n}=-7\hat{i}-7\hat{j}+7\hat{k}$

Thus, the direction ratios of the normal to the plane are $<-7, -7, 7>.$

The equation of a plane having direction ratios of the normal to the plane as $a, b, c$ and passing through a point $\left(x_1, y_1, z_1\right)$ is $a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$

Hence, the equation of the plane having direction ratios the normal to the plane as $<-7, -7, 7>$ and passing through the point$\left(4, -1, 2\right)$ is 

$-7\left(x-4\right)-7\left(y+1\right)+7\left(z-2\right)=0$

$\Rightarrow -7x-7y+7z=-7$

$\Rightarrow x+y-z=1$

From the given options only the point $\left(1, 1, 1\right)$ satisfy the equation of the plane.

Hence, the plane passes through the point $\left(1, 1, 1\right).$