The problem asks to find the symmetric equations of the tangent line to the curve given by the vector function $\mathbf{r}(t) = 2\cos{t} \mathbf{i} + 6\sin{t} \mathbf{j} + t \mathbf{k}$ at $t = \pi/3$.

1. Problem Description

The problem asks to find the symmetric equations of the tangent line to the curve given by the vector function $\mathbf{r}(t) = 2\cos{t} \mathbf{i} + 6\sin{t} \mathbf{j} + t \mathbf{k}$ at $t = \pi/3$.

2. Solution Steps

First, find the derivative of $\mathbf{r}(t)$ with respect to $t$:

$\mathbf{r}'(t) = -2\sin{t} \mathbf{i} + 6\cos{t} \mathbf{j} + \mathbf{k}$

Next, evaluate $\mathbf{r}'(t)$ at $t = \pi/3$:

$\mathbf{r}'(\pi/3) = -2\sin{(\pi/3)} \mathbf{i} + 6\cos{(\pi/3)} \mathbf{j} + \mathbf{k} = -2(\sqrt{3}/2) \mathbf{i} + 6(1/2) \mathbf{j} + \mathbf{k} = -\sqrt{3} \mathbf{i} + 3 \mathbf{j} + \mathbf{k}$

This vector $\mathbf{r}'(\pi/3) = \langle -\sqrt{3}, 3, 1 \rangle$ gives the direction vector of the tangent line.

Now, evaluate $\mathbf{r}(t)$ at $t = \pi/3$:

$\mathbf{r}(\pi/3) = 2\cos{(\pi/3)} \mathbf{i} + 6\sin{(\pi/3)} \mathbf{j} + (\pi/3) \mathbf{k} = 2(1/2) \mathbf{i} + 6(\sqrt{3}/2) \mathbf{j} + (\pi/3) \mathbf{k} = \mathbf{i} + 3\sqrt{3} \mathbf{j} + (\pi/3) \mathbf{k}$

So the point on the curve at $t = \pi/3$ is $(1, 3\sqrt{3}, \pi/3)$.

The parametric equations of the tangent line are:

$x = 1 - \sqrt{3}t$

$y = 3\sqrt{3} + 3t$

$z = \pi/3 + t$

To find the symmetric equations, solve for $t$ in each equation:

$t = \frac{1-x}{\sqrt{3}}$

$t = \frac{y - 3\sqrt{3}}{3}$

$t = z - \frac{\pi}{3}$

Now, set these expressions for $t$ equal to each other:

$\frac{1-x}{\sqrt{3}} = \frac{y - 3\sqrt{3}}{3} = z - \frac{\pi}{3}$

3. Final Answer

The symmetric equations of the tangent line are:

$\frac{1-x}{\sqrt{3}} = \frac{y - 3\sqrt{3}}{3} = z - \frac{\pi}{3}$