Let point $Q$ is $\left(x_1,\mathrm{}{y}_1,\mathrm{}{z}_1\right)$
D.R’s of $PQ$ are $x_1-1,\mathrm{}{y}_1-1,\mathrm{}{z}_1-1$
$\Rightarrow 1\left(x_1-1\right)+1\left(y_1-1\right)+1\left(z_1-1\right)=0$
$\Rightarrow x_1+y_1+z_1=3$
Also $x_1+2y_1+3z_1=4$
$x_1+y_1+z_1=3$
$\Rightarrow \left(x_1,y_1,\mathrm{}{z}_1\right)$ is a point on the two planes
$x+2y+3z=4$
$x+y+z=3$
$\therefore$ the line on the intersection of these two planes will be the required locus

Now, the direction vector of the line is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 2& 3\end{array}\right|=\hat{i}-2\hat{j}+\hat{k}$

Also, the point $\left(0,5,-2\right)$ satisfy both the equation of planes

Hence, the equation of the required line is $\frac{x}{1}=\frac{y-5}{-2}=\frac{z+2}{1}$