Let $$\vec a = \hat j - \hat k$$   and $$\vec c = \hat i - \hat j - \,\hat k.$$   Then the vector $${\vec b}$$ satisfying $$\vec a \times \vec b + \vec c = 0$$   and $$\vec a.\vec b = 3$$

Solution :
$$\eqalign{ & \vec c = \vec b \times \vec a\,\, \Rightarrow \vec b.\vec c = \vec b.\left( {\vec b \times \vec a} \right)\,\, \Rightarrow \vec b.\vec c = 0 \cr & \Rightarrow \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right).\left( {\hat i - \hat j - \hat k} \right) = 0, \cr & {\text{where }}\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k \cr & {b_1} + {b_2} + {b_3} = 0.....({\text{i}}) \cr & {\text{and }}\vec a.\vec b = 3\,\, \Rightarrow \left( {\hat j - \hat k} \right).\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = 3 \cr & \Rightarrow {b_2} - {b_3} = 3 \cr} $$
From equation (i)
$$\eqalign{ & {b_1} = {b_2} + {b_3} = \left( {3 + {b_3}} \right) + {b_3} = 3 + 2{b_3} \cr & \vec b = \left( {3 + 2{b_3}} \right)\hat i + \left( {3 + {b_3}} \right)\hat j + {b_3}\hat k \cr} $$
From the option given, it is clear that $${b_3}$$ equal to either 2 or $$-2$$
If $${b_3} = 2,$$   then $$\vec b = 7\hat i + 5\hat j + 2\hat k$$    which is not possible
If $${b_3} = - 2,$$   then $$\vec b = - \hat i + \hat j - 2\hat k$$