If $\vec r$ satisfies the equation $\vec r× (\hat i+2\hat j+\hat k)=\hat i-\hat k$, then for any scalar m, $\vec r$ is equal to

$\hat j+m(\hat i + 2\hat j+\hat k)$

Let $\vec r=x\hat i+y\hat j+z\hat k$. Then,

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

$⇒(y-2z)\hat i+ (z - x)\hat j + (2x-y)\hat k=\hat i-\hat k$

$⇒y-2z=1,z-x=0$ and $2x-y=-1$

$⇒\frac{x}{1}=\frac{z}{1}=\frac{y-1}{2}=m (say)$

Then, $x =m, z=m$ and $y = 2m+1$

$∴\vec r=\hat j+m(\hat i + 2\hat j+\hat k)$