Step-by-step Derivations:

1. Express vector v in terms of a and b

The problem states that vector $\overrightarrow v$ lies in the plane of vectors $\overrightarrow a$ and $\overrightarrow b$. This means $\overrightarrow v$ can be expressed as a linear combination of $\overrightarrow a$ and $\overrightarrow b$. Let $\overrightarrow v = \lambda \overrightarrow a + \mu \overrightarrow b$ for some scalars $\lambda$ and $\mu$.

Given: $\overrightarrow a = \widehat i + \widehat j + \widehat k$ $\overrightarrow b = \widehat i - \widehat j + \widehat k$

Substituting these into the expression for $\overrightarrow v$: $\overrightarrow v = \lambda (\widehat i + \widehat j + \widehat k) + \mu (\widehat i - \widehat j + \widehat k)$ $\overrightarrow v = (\lambda + \mu)\widehat i + (\lambda - \mu)\widehat j + (\lambda + \mu)\widehat k$

2. Use the projection condition

The projection of a vector $\overrightarrow v$ onto a vector $\overrightarrow c$ is given by the formula: $Projection = \frac{\overrightarrow v \cdot \overrightarrow c}{|\overrightarrow c|}$ We are given that the projection of $\overrightarrow v$ on $\overrightarrow c$ is $\frac{1}{\sqrt{3}}$.

Given $\overrightarrow c = \widehat i - \widehat j - \widehat k$, let's first calculate its magnitude $|\overrightarrow c|$: $|\overrightarrow c| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$

Now, let's calculate the dot product $\overrightarrow v \cdot \overrightarrow c$: $\overrightarrow v \cdot \overrightarrow c = [(\lambda + \mu)\widehat i + (\lambda - \mu)\widehat j + (\lambda + \mu)\widehat k] \cdot [\widehat i - \widehat j - \widehat k]$ $\overrightarrow v \cdot \overrightarrow c = (\lambda + \mu)(1) + (\lambda - \mu)(-1) + (\lambda + \mu)(-1)$ $\overrightarrow v \cdot \overrightarrow c = (\lambda + \mu) - (\lambda - \mu) - (\lambda + \mu)$ $\overrightarrow v \cdot \overrightarrow c = \lambda + \mu - \lambda + \mu - \lambda - \mu = \mu - \lambda$

Now we can use the projection equation: $\frac{\overrightarrow v \cdot \overrightarrow c}{|\overrightarrow c|} = \frac{\mu - \lambda}{\sqrt{3}} = \frac{1}{\sqrt{3}}$ This implies $\mu - \lambda = 1$, or $\mu = \lambda + 1$.

3. Substitute to find the general form of v

Now we substitute $\mu = \lambda + 1$ back into our expression for $\overrightarrow v$: $\overrightarrow v = (\lambda + (\lambda+1))\widehat i + (\lambda - (\lambda+1))\widehat j + (\lambda + (\lambda+1))\widehat k$ $\overrightarrow v = (2\lambda + 1)\widehat i - 1\widehat j + (2\lambda + 1)\widehat k$

4. Match the general form with the given options

The general form of the vector $\overrightarrow v$ is $(2\lambda + 1)\widehat i - \widehat j + (2\lambda + 1)\widehat k$. We can see that the $\widehat j$ component of $\overrightarrow v$ must be $-1$. Let's check the options:

A: $\widehat i - 3\widehat j + 3\widehat k$ (The $\widehat j$ component is -3. Incorrect.) B: $-3\widehat i - 3\widehat j - \widehat k$ (The $\widehat j$ component is -3. Incorrect.) C: $3\widehat i - \widehat j + 3\widehat k$ (The $\widehat j$ component is -1. This is a possible answer.) D: $\widehat i + 3\widehat j - 3\widehat k$ (The $\widehat j$ component is 3. Incorrect.)

Only option C has a $\widehat j$ component of $-1$. Let's verify if the other components match for some value of $\lambda$. Comparing $3\widehat i - \widehat j + 3\widehat k$ with $(2\lambda + 1)\widehat i - \widehat j + (2\lambda + 1)\widehat k$, we must have: $2\lambda + 1 = 3$ $2\lambda = 2$ $\lambda = 1$ This gives a consistent value for $\lambda$. If $\lambda=1$, then $\mu = \lambda + 1 = 2$. The vector $\overrightarrow v$ is $1 \cdot \overrightarrow a + 2 \cdot \overrightarrow b = (\widehat i + \widehat j + \widehat k) + 2(\widehat i - \widehat j + \widehat k) = (1+2)\widehat i + (1-2)\widehat j + (1+2)\widehat k = 3\widehat i - \widehat j + 3\widehat k$ This matches option C perfectly.

Conclusion:

The vector $\overrightarrow v$ is $3\widehat i - \widehat j + 3\widehat k$.