If vectors $$\vec{a_1} = x\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{a_2} = \hat{i} + y\hat{j} + z\hat{k}$$ are collinear, then a possible unit vector parallel to the vector $$x\hat{i} + y\hat{j} + z\hat{k}$$ is:

Since $$\vec{a_1} = x\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{a_2} = \hat{i} + y\hat{j} + z\hat{k}$$ are collinear, there exists a scalar $$\lambda$$ such that $$\vec{a_2} = \lambda \vec{a_1}$$. Writing this component-wise: $$\hat{i} + y\hat{j} + z\hat{k} = \lambda(x\hat{i} - \hat{j} + \hat{k})$$.

Comparing the $$\hat{i}$$ component: $$1 = \lambda x$$, so $$\lambda = \frac{1}{x}$$.

Comparing the $$\hat{j}$$ component: $$y = -\lambda = -\frac{1}{x}$$.

Comparing the $$\hat{k}$$ component: $$z = \lambda = \frac{1}{x}$$.

Now we form the vector $$x\hat{i} + y\hat{j} + z\hat{k} = x\hat{i} - \frac{1}{x}\hat{j} + \frac{1}{x}\hat{k}$$. To find a unit vector parallel to this, we can choose a convenient value of $$x$$. Since $$x$$ can be any nonzero real number (the collinearity condition allows any $$x \neq 0$$), let us try $$x = 1$$.

With $$x = 1$$: $$y = -\frac{1}{1} = -1$$ and $$z = \frac{1}{1} = 1$$. We verify collinearity: $$\vec{a_1} = \hat{i} - \hat{j} + \hat{k}$$ and $$\vec{a_2} = \hat{i} - \hat{j} + \hat{k}$$, which are indeed collinear ($$\lambda = 1$$).

The vector $$x\hat{i} + y\hat{j} + z\hat{k} = \hat{i} - \hat{j} + \hat{k}$$, with magnitude $$|\hat{i} - \hat{j} + \hat{k}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$.

The unit vector is $$\frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$$.