1. Let $\vec a_1 = x\hat i - \hat j + \hat k = (x,-1,1), \qquad \vec a_2 = \hat i + y\hat j + z\hat k = (1,y,z).$

Since the vectors are collinear, one must be a scalar multiple of the other. So there exists some scalar $\lambda$ such that $\vec a_1 = \lambda \vec a_2.$

Thus, $(x,-1,1)=\lambda(1,y,z).$

2. Compare components: $x=\lambda, \qquad -1=\lambda y, \qquad 1=\lambda z.$

From these, $y=\frac{-1}{\lambda}, \qquad z=\frac{1}{\lambda}.$ Since $x=\lambda$, we get $y=-\frac{1}{x}, \qquad z=\frac{1}{x}.$

3. Now consider the vector $x\hat i + y\hat j + z\hat k = \left(x,-\frac1x,\frac1x\right).$

Using $x=\lambda$, this becomes $\left(\lambda,-\frac1\lambda,\frac1\lambda\right).$

To determine its direction, use the collinearity relation more directly. From $\frac{x}{1}=\frac{-1}{y}=\frac{1}{z}=\lambda,$ we get $xy=-1, \qquad xz=1.$ Hence $y=-\frac1x, \qquad z=\frac1x.$ So the vector is $x\hat i+y\hat j+z\hat k = x\hat i-\frac1x\hat j+\frac1x\hat k.$

For the direction to match one of the options, check whether this vector can be parallel to $\hat i-\hat j+\hat k.$ Indeed, if $x=1$, then $(x,y,z)=(1,-1,1),$ and the vector becomes $\hat i-\hat j+\hat k.$ Its unit vector is $\frac{1}{\sqrt{1^2+(-1)^2+1^2}}(\hat i-\hat j+\hat k)=\frac{1}{\sqrt3}(\hat i-\hat j+\hat k).$

4. Check options:

• Option A: $\frac{1}{\sqrt3}(\hat i-\hat j+\hat k)$ ✓
• Option B has no $\hat i$ component, so not parallel.
• Option C has no $\hat k$ component, so not parallel.
• Option D has signs different from the required direction.

Therefore, the possible unit vector is $\boxed{\frac{1}{\sqrt3}(\hat i-\hat j+\hat k)}.$