Answer
$xy$ plane
Work Step by Step
Given
The initial position vector of the proton $\overrightarrow{r_1}= 5\hat{i}-6\hat{j}+2\hat{k}$
The final position vector of the proton $\overrightarrow{r_2} = -2\hat{i}+6\hat{j}+2\hat{k}$
The displacement vector $\overrightarrow{d} =\overrightarrow{r_2}-\overrightarrow{r_1}$
$\overrightarrow{d}=-2\hat{i}+6\hat{j}+2\hat{k} - (5\hat{i}-6\hat{j}+2\hat{k})$
$\overrightarrow{d} =-7\hat{i}+12\hat{j}+0\hat{k} $
since In displacement vector $\hat{k}$ component is zero or not present.so the vector $\overrightarrow{d}$ must lie in xy plane.i,e $\overrightarrow{d}$ is parallel to the xy plane.
