If $\vec a =\hat i +\hat j+\hat k$ and $\vec b =\hat i-2\hat j+\hat k$, then the vector $\vec c$ such that $\vec a.\vec c = 2$ and $\vec a ×\vec c =\vec b$ is

$\frac{1}{3}(-\hat i+2\hat j+5\hat k)$

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

$\vec a.\vec c = 2$ and $\vec a ×\vec c =\vec b$

$⇒\vec a.\vec c = 2$ and $(\vec a ×\vec c).\vec c=\vec b.\vec c$

$⇒\vec a.\vec c = 2$ and $\vec b.\vec c=0$   $[∵(\vec a ×\vec c).\vec c=0]$

$⇒x+y+z=2$ and $x-2y+z=0$

$⇒y=\frac{2}{3}$ and $x+z=\frac{4}{3}$

Clearly, option (2) satisfies the above conditions.