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

The given vectors are
$$\vec{a}=\,\hat{j}-\hat{k}=(0,\,1,\,-1),\qquad \vec{c}=\,\hat{i}-\hat{j}-\hat{k}=(1,\,-1,\,-1).$$

Let $$\vec{b}=x\,\hat{i}+y\,\hat{j}+z\,\hat{k}=(x,\,y,\,z).$$

Step 1: Use $$\vec{a}\times\vec{b}+\vec{c}=\,\vec{0}$$.
Compute the cross product with the determinant formula:

$$ \vec{a}\times\vec{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 0&1&-1\\ x&y&z \end{vmatrix} =\hat{i}(1\cdot z-(-1)\cdot y)-\hat{j}(0\cdot z-(-1)\cdot x)+\hat{k}(0\cdot y-1\cdot x).$$

Simplify each component:

$$ \vec{a}\times\vec{b}=(\,z+y,\,-x,\,-x). $$

The condition $$\vec{a}\times\vec{b}+\vec{c}=\,\vec{0}$$ is therefore

$$ (z+y,\,-x,\,-x)+(1,\,-1,\,-1)=(0,0,0). $$

Equating components gives the first set of equations:

$$\begin{aligned} z+y &= -1, \quad\;(i)\\ -x &= 1 \;\Rightarrow\; x=-1, \quad\;(ii)\\ -x &= 1 \;\;\text{(same as above).} \end{aligned}$$

Step 2: Use the dot-product condition $$\vec{a}\cdot\vec{b}=3$$.
Since $$\vec{a}=(0,1,-1),$$

$$ \vec{a}\cdot\vec{b}=0\cdot x+1\cdot y+(-1)\cdot z=y-z=3. \quad\;(iii) $$

Step 3: Solve for $$y$$ and $$z$$.
From (i) and (iii):

$$\begin{aligned} y+z &= -1,\\ y-z &= 3. \end{aligned}$$

Add the two equations to get $$2y=2 \Rightarrow y=1.$$
Substitute in $$y-z=3$$ to get $$z=y-3=1-3=-2.$$

Step 4: Form the required vector.
With $$x=-1,\;y=1,\;z=-2,$$ we obtain

$$ \vec{b}=-\hat{i}+\hat{j}-2\hat{k}. $$

Hence the vector $$\vec{b}$$ is $$-\hat{i}+\hat{j}-2\hat{k}$$.

Option D which is: $$-\hat{i} + \hat{j} - 2\hat{k}$$