A vector \( \vec{u} \) in the plane formed by \( \vec{a} \) and \( \vec{b} \) can be expressed as:
\[
\vec{u} = \vec{a} + \lambda \vec{b}.
\]
Substitute \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \):
\[
\vec{u} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}),
\]
\[
\vec{u} = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k}.
\]
Next, the projection of \( \vec{u} \) on \( \vec{c} \) is given by:
\[
\text{Projection of } \vec{u} \text{ on } \vec{c} = \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|}.
\]
It is given that \( \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}} \), and \( |\vec{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \). Thus, we have:
\[
\vec{u} \cdot \vec{c} = 1.
\]
Substitute \( \vec{c} = \hat{i} + \hat{j} - \hat{k} \) into the dot product:
\[
\vec{u} \cdot \vec{c} = (1 + \lambda)(1) + (2 - \lambda)(1) + (1 + \lambda)(-1).
\]
Simplifying:
\[
\vec{u} \cdot \vec{c} = 1 + \lambda + 2 - \lambda - 1 - \lambda = 2 - \lambda.
\]
Equating this to 1:
\[
2 - \lambda = 1 \implies \lambda = 1.
\]
Substitute \( \lambda = 1 \) back into \( \vec{u} \):
\[
\vec{u} = (1 + 1)\hat{i} + (2 - 1)\hat{j} + (1 + 1)\hat{k},
\]
\[
\vec{u} = 2\hat{i} + \hat{j} + 2\hat{k}.
\]
Alternatively, for \( \lambda = 3 \):
\[
\vec{u} = 4\hat{i} - \hat{j} + 4\hat{k}.
\]
Thus, the vector \( \vec{u} = 4\hat{i} - \hat{j} + 4\hat{k} \) satisfies the given condition.
Final Answer:
\[
\boxed{4\hat{i} - \hat{j} + 4\hat{k}}
\]