Let $\vec a =\hat i +\hat j+\hat k, \vec c=\hat j-\hat k$. If $\vec b$ is a vector satisfying $\vec a × \vec b = \vec c$ and $\vec a. \vec b = 3$, then $\vec b$ is

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

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

$\vec a.\vec b=3⇒x+y+z=3$

and,

$\vec a × \vec b = \vec c$

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

$⇒z-y=0,x-z=1$ and $y-x=-1$

Solving these equations with $x + y + z = 3$, we get

$x=5/3, y=2/3$ and $z = 2/3$

$∴\vec b=\frac{1}{3}(5\hat i+2\hat j+2\hat k)$