If $$\vec{a}$$ is nonzero vector such that its projections on the vectors $$2\hat{i} - \hat{j} + 2\hat{k}$$, $$\hat{i} + 2\hat{j} - 2\hat{k}$$ and $$\hat{k}$$ are equal, then a unit vector along $$\vec{a}$$ is:

Let $$\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$$ be a nonzero vector. The projection of $$\vec{a}$$ on a vector $$\vec{v}$$ is $$\frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$$.

The three given vectors are $$\vec{v_1} = 2\hat{i} - \hat{j} + 2\hat{k}$$, $$\vec{v_2} = \hat{i} + 2\hat{j} - 2\hat{k}$$, and $$\vec{v_3} = \hat{k}$$.

Their magnitudes are: $$|\vec{v_1}| = \sqrt{4+1+4} = 3$$, $$|\vec{v_2}| = \sqrt{1+4+4} = 3$$, $$|\vec{v_3}| = 1$$.

The projections are:

$$\text{proj}_1 = \frac{2x - y + 2z}{3}, \quad \text{proj}_2 = \frac{x + 2y - 2z}{3}, \quad \text{proj}_3 = z$$

Setting proj$$_1$$ = proj$$_2$$: $$2x - y + 2z = x + 2y - 2z$$, which gives $$x - 3y + 4z = 0$$ ...(1)

Setting proj$$_2$$ = proj$$_3$$: $$\frac{x + 2y - 2z}{3} = z$$, which gives $$x + 2y - 5z = 0$$ ...(2)

From (1) and (2): subtracting (2) from (1): $$-5y + 9z = 0$$, so $$y = \frac{9z}{5}$$.

Substituting into (2): $$x + \frac{18z}{5} - 5z = 0$$, so $$x = 5z - \frac{18z}{5} = \frac{7z}{5}$$.

Taking $$z = 5$$: $$\vec{a} = 7\hat{i} + 9\hat{j} + 5\hat{k}$$.

$$|\vec{a}| = \sqrt{49 + 81 + 25} = \sqrt{155}$$.

The unit vector along $$\vec{a}$$ is $$\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$$.

Hence, the correct answer is Option C.