Let $$\vec{A} = (\hat{i} + \hat{j})$$ and $$\vec{B} = (2\hat{i} - \hat{j})$$. The magnitude of a coplanar vector $$\vec{C}$$ such that $$\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C} = \vec{A} \cdot \vec{B}$$ is given by:

We are given two vectors in the plane

$$\vec A = (\,\hat i + \hat j\,) \qquad\text{and}\qquad \vec B = (\,2\hat i - \hat j\,).$$

Let the required coplanar vector be

$$\vec C = (\,x\hat i + y\hat j\,).$$

Because all three vectors lie in the same plane, no further restriction on direction is needed; any ordered pair $$\,(x,y)\,$ represents a coplanar vector.

The condition stated in the problem is

$$\vec A\cdot\vec C \;=\;\vec B\cdot\vec C \;=\;\vec A\cdot\vec B.$$

First we compute $$\vec A\cdot\vec B.$$ The dot-product formula is

$$\vec u\cdot\vec v = u_xv_x + u_yv_y,$$ so, substituting $$\vec A = (1,1)$$ and $$\vec B = (2,-1),$$ we obtain

$$\vec A\cdot\vec B \;=\; 1\cdot 2 + 1\cdot(-1) \;=\; 2 - 1 \;=\; 1.$$

Next we impose the two equality conditions one by one.

1. The dot product $$\vec A\cdot\vec C$$ is, again using the same formula,

$$\vec A\cdot\vec C \;=\; 1\cdot x + 1\cdot y \;=\; x + y.$$

2. Similarly,

$$\vec B\cdot\vec C \;=\; 2\cdot x + (-1)\cdot y \;=\; 2x - y.$$

The given relations now read

$$x + y = 1 \quad\text{and}\quad 2x - y = 1.$$

We solve this pair of simultaneous linear equations step by step. Adding the two equations eliminates $$y$$:

$$(x + y) + (2x - y) \;=\; 1 + 1 \;\;\Longrightarrow\;\; 3x \;=\; 2 \;\;\Longrightarrow\;\; x \;=\; \frac{2}{3}.$$

Substituting $$x = \dfrac{2}{3}$$ back into $$x + y = 1$$ gives

$$\frac{2}{3} + y = 1 \;\;\Longrightarrow\;\; y = 1 - \frac{2}{3} = \frac{1}{3}.$$

Hence the vector $$\vec C$$ that satisfies both equalities is

$$\vec C = \left(\frac{2}{3}\hat i + \frac{1}{3}\hat j\right).$$

Finally, we calculate its magnitude. The magnitude (length) of a vector $$\vec v = (v_x,v_y)$$ is given by

$$|\vec v| = \sqrt{v_x^{\,2} + v_y^{\,2}}.$$

Applying this to $$\vec C,$$ we have

$$|\vec C| = \sqrt{\left(\frac{2}{3}\right)^{\!2} + \left(\frac{1}{3}\right)^{\!2}} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt5}{3}.$$

Thus the magnitude of the required vector is $$\sqrt{\dfrac{5}{9}}.$$

Hence, the correct answer is Option C.