$\mathbf{a} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} + \hat{\mathbf{k}}$

$\mathbf{a} \cdot \hat{\mathbf{i}} = 1$

First, find the vector $\mathbf{a}$. Assume $\mathbf{a}$ can be expressed as $x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$. We have:

$$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \times \hat{\mathbf{i}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $$

Breaking down the cross product:

$x \hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0$

$y \hat{\mathbf{j}} \times \hat{\mathbf{i}} = -y \hat{\mathbf{k}}$

$z \hat{\mathbf{k}} \times \hat{\mathbf{i}} = z \hat{\mathbf{j}}$

Combining, we get:

$$ -y \hat{\mathbf{k}} + z \hat{\mathbf{j}} = \hat{\mathbf{j}} + \hat{\mathbf{k}} $$

Equating components:

$z = 1$

$-y = 1 \Rightarrow y = -1$

Given $\mathbf{a} \cdot \hat{\mathbf{i}} = 1$:

$$ (x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}) \cdot \hat{\mathbf{i}} = x = 1 $$

Thus, we find:

$$ \mathbf{a} = \hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}} $$

Now, the equation of the line passing through the point $\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}$ and parallel to the vector $\mathbf{a}$ is:

$$ \mathbf{r} = (\text{Passing through point}) + t (\text{Parallel vector}) $$

Substitute:

$$ \mathbf{r} = (\hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}}) + t (\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}) $$

Simplify:

$$ \mathbf{r} = (1 + t) \hat{\mathbf{i}} + (1 - t) \hat{\mathbf{j}} + (1 + t) \hat{\mathbf{k}} $$

This is the equation of the line.

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