A unit vector perpendicular to the vector $-\hat i+2\hat j+2\hat k$ and making equal angles with X and Y axes can be

$\frac{1}{3}(2\hat i+2\hat j-\hat k)$

Let the required vector be $\vec r=l\hat i+m\hat j+n\hat k$ where $l, m, n$ are the direction cosines of $\vec r$ such that $l = m$.

It is given that $\vec r$ is perpendicular to $-\hat i + 2\hat + 2\hat k$. Therefore,

$\vec r.(-\hat i+2\hat j+2\hat k) =0$

$⇒-l+2m+2n=0$

$⇒l+2n=0$   $[∵l=m]$

$⇒l=-2n$

Now,

$l^2+ m^2 + n^2=1⇒ 4n^2 + 4n^2+n^2=1⇒n=±\frac{1}{3}$

$∴l=±\frac{2}{3},m=±\frac{2}{3}$ and $n=±\frac{1}{3}$

Hence, $\vec r=±\frac{1}{3}(2\hat i+2\hat j-\hat k)$