Let (\vec{a} = \hat{j} - \hat{k}) and (\vec{c} = \hat{i} - \hat{j} - \hat{k}). Then vector (\vec{b}) satisfying (\vec{a} \times \vec{b} + \vec{c} = \vec{0}) and (\vec{a} \cdot \vec{b} = 3) is

Given that (\vec{a} \times \vec{b} + \vec{c} = \vec{0} \implies \vec{c} = \vec{b} \times \vec{a}
So we get (\vec{b} \cdot \vec{c} = 0) (1)
Let (\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}
Putting values in Eq-1 ((b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = 0
b_1 - b_2 - b_3 = 0
Now given that (\vec{a} \cdot \vec{b} = 3 = b_2 - b_3 = 3
b_2 = 3 + b_3
b_1 = b_2 + b_3 = 3 + 2b_3
\vec{b} = (3 + 2b_3)\hat{i} + (3 + b_3)\hat{j} + b_3\hat{k}