vector.
since $\hat{a}$ is perpendicular to $(2 \hat{i}-\hat{j}+2 \hat{k})$
$\therefore \quad 2 x-y+2 z=0$ ......$(i)$
Since vecter $x\hat i + y\hat j + z\hat k$ is coplanar with the vector $\hat i + \hat j - \hat k$ and $2 \hat{i}+2 \hat{j}-\hat{k}$
$\therefore \quad x \hat{i}+y \hat{j}+z \hat{k}$
$ = p(\hat i + \hat j - \hat k) + q(2\hat i + 2\hat j - \hat k),$
where $p$ and $q$ are some scalars.
$ \Rightarrow \quad x\hat i + y\hat j + x\hat k$
$ = (p + 2q)\hat i + (p + 2q)\hat j - (p + q)\hat k$
$ \Rightarrow \quad x = p + 2q,y = p + 2q,z = - p - q$
Now from equation $ ( i)$
$2 p+4 q-p-2 q-2 p-2 q=0$
$\Rightarrow \quad-p=0 \Rightarrow p=0$
$\therefore \quad x = 2q,y - 2q,z = - q$
Since vect cr $x \hat{i}+y \hat{j}+z \hat{k}$ is a unit vect $\sigma$ therefice
$|x\hat i + y\hat j + z\hat k| = 1$
$ \Rightarrow \quad \sqrt {{x^2} + {y^2} + {z^2}} =1$
$ \Rightarrow \quad {x^2} + {y^2} + {z^2} = 1$
$ \Rightarrow \quad 4{q^2} + 4{q^2} + {q^2} = 1$
$ \Rightarrow 9{q^2} - 1 \Rightarrow q = \pm \frac{1}{3}$
When $q=\frac{1}{3},$ then $x=\frac{2}{3}, y=\frac{2}{3}$
$z=-\frac{1}{3}$
When $q = - \frac{1}{3},$ then $x=-\frac{2}{3}, y=-\frac{2}{3}$
$z=\frac{1}{3}$
Hare required unit vector is $\frac{2}{3}\hat i + \frac{2}{3}\hat j - \frac{1}{3}\hat k$
or $ - \frac{2}{3}\hat i - \frac{2}{3}\hat j + \frac{1}{3}\hat k$.
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