To find the point at which the line intersects the $xz$-plane, we need to set up the condition for this intersection.

The equation of the line is given by: $\vec{r}(t) = \langle -4, -6, 6 \rangle + t \langle -8, -1, -2 \rangle$ where $t$ is a parameter.

The $xz$-plane is defined by $y = 0$. So, to find the point of intersection with the $xz$-plane, we need to set the $y$-component of $\vec{r}(t)$ to zero and solve for $t$.

The $y$-component of $\vec{r}(t)$ is: $-6 + t(-1) = -6 - t$ Setting this equal to zero: $-6 - t = 0$ $t = -6$

Now, substitute $t = -6$ into $\vec{r}(t)$ to find the coordinates of the point of intersection.

Calculating $x$-component:

$x = -4 + (-6)(-8) = -4 + 48 = 44$

Calculating $y$-component:

We set $y = 0$ by solving for $t$, so this is already satisfied.

Calculating $z$-component:

$z = 6 + (-6)(-2) = 6 + 12 = 18$

Therefore, the point at which the line intersects the $xz$-plane is: $(44, 0, 18)$

Would you like a more detailed explanation or have any questions?

Here are some related questions you might find helpful:

1. How do you find the point of intersection of a line with a different plane?
2. What if the line did not intersect the plane? How would you identify that?
3. How can parametric equations of a line be used to find intersections with planes in 3D?
4. What is the significance of different planes in 3D coordinate geometry?
5. How would you convert parametric equations into symmetric form?

Tip: Remember that the condition for a line to intersect the $xz$-plane is simply that the $y$-component becomes zero. This concept can be extended to find intersections with any plane by setting the respective component to zero or a specific value.