Step 1: Set up the unknown vector b.
Let $\vec{b} = x\hat{i}+y\hat{j}+z\hat{k}$. We are given $\vec{a} = \hat{i}+\hat{j}+\hat{k}$, $\vec{c} = \hat{j}-\hat{k}$, $\vec{a}\times\vec{b}=\vec{c}$, and $\vec{a}\cdot\vec{b}=3$.
Step 2: Use the dot product condition to get equation 1.
$\vec{a}\cdot\vec{b} = (1)(x)+(1)(y)+(1)(z) = x+y+z = 3$. ... (i)
Step 3: Compute the cross product a x b.
\[ \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&1\\x&y&z\end{vmatrix} = (z-y)\hat{i}-(z-x)\hat{j}+(y-x)\hat{k}. \] This must equal $\vec{c} = 0\hat{i}+1\hat{j}-1\hat{k}$.
Step 4: Match components to get three more equations.
Comparing $\hat{i}$: $z-y=0 \Rightarrow y=z$. ... (ii)
Comparing $\hat{j}$: $-(z-x)=1 \Rightarrow x-z=1$. ... (iii)
Comparing $\hat{k}$: $y-x=-1 \Rightarrow x-y=1$. ... (iv)
Step 5: Solve the system of equations.
From (ii): $y=z$. From (iii): $x = z+1$. Substitute into (i): $(z+1)+z+z = 3 \Rightarrow 3z+1=3 \Rightarrow z = \frac{2}{3}$. Then $y = \frac{2}{3}$ and $x = \frac{2}{3}+1 = \frac{5}{3}$. So $\vec{b} = \frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})$.
Step 6: Verify and state the answer.
Check: $\vec{a}\cdot\vec{b} = \frac{5}{3}+\frac{2}{3}+\frac{2}{3} = \frac{9}{3} = 3$. Check $\vec{a}\times\vec{b}$: $z-y=0$, $-(z-x)=-(\frac{2}{3}-\frac{5}{3})=1$, $y-x=\frac{2}{3}-\frac{5}{3}=-1$. So $\vec{a}\times\vec{b}=0\hat{i}+1\hat{j}-1\hat{k}=\hat{j}-\hat{k}=\vec{c}$. Correct. \[ \boxed{\vec{b} = \dfrac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})} \]