The diagram below shows a tetrahedron $ABCD$. The equation of the plane $ABD$ is $4x + y + 2z = 16$.

1. Use the property that points on the $x$-axis have $y=0$ and $z=0$ to find the $x$-coordinate.

Since $A$ is on the $x$-axis, its $y$ and $z$ coordinates are $0$.
Substitute $y=0$ and $z=0$ into the equation of the plane $ABD$ ($4x + y + 2z = 16$):
$4x + 0 + 2(0) = 16$
$4x = 16$
$x = 4$

2. State the coordinates of $A$.

The coordinates of $A$ are $(4, 0, 0)$.

1. Identify the relationship between the required line and the two given planes, and find its direction vector.

The points $B$ and $D$ lie on both the plane $ABD$ and the plane $CBD$. Thus, the line passing through $B$ and $D$ is the line of intersection of these two planes.
Its direction vector $\mathbf{d}$ must be perpendicular to the normal vectors of both planes. We can find this using the cross product:
$\mathbf{d} = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 7 \\ -11 \\ -5 \end{pmatrix} = \begin{pmatrix} (1)(-5) - (2)(-11) \\ (2)(7) - (4)(-5) \\ (4)(-11) - (1)(7) \end{pmatrix} = \begin{pmatrix} -5 + 22 \\ 14 + 20 \\ -44 - 7 \end{pmatrix} = \begin{pmatrix} 17 \\ 34 \\ -51 \end{pmatrix} = 17\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$
We can take the simplified direction vector to be $\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$.

2. Find a specific point on the line by solving the plane equations simultaneously.

To find a point on the line of intersection, we can substitute a value for one of the coordinates into the equations of the planes. Let $z = 0$:
$4x + y = 16$
$7x - 11y = -23$
From the first equation, $y = 16 - 4x$. Substituting this into the second equation gives:
$7x - 11(16 - 4x) = -23$
$7x - 176 + 44x = -23$
$51x = 153 \implies x = 3$
Substituting $x = 3$ back to find $y$:
$y = 16 - 4(3) = 4$
A point on the line is $(3, 4, 0)$.

3. Formulate the final vector equation of the line.

Using the point $(3, 4, 0)$ and the direction vector $\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$, the vector equation of the line is:
$\mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

1. Formulate a system of equations using the fact that $B$ lies on the given planes and the $xy$-plane.

Since $B$ is on the $xy$-plane, its $z$-coordinate is $0$. Let $B$ be $(x, y, 0)$.
Point $B$ must lie on both the plane $ABD$ and the plane $CBD$. Substituting $z = 0$ into their respective equations gives:
$4x + y = 16 \quad \text{--- (1)}$
$7x - 11y = -23 \quad \text{--- (2)}$

2. Solve the simultaneous equations to find the coordinates of $B$.

From (1), we have $y = 16 - 4x$.
Substitute $y$ into (2):
$7x - 11(16 - 4x) = -23$
$7x - 176 + 44x = -23$
$51x = 153$
$x = 3$
Substitute $x = 3$ back into the equation for $y$:
$y = 16 - 4(3) = 4$
Thus, the coordinates of $B$ are $(3, 4, 0)$.

1. Express the given line $AD$ in parametric form.

Let $\frac{4-x}{2} = \frac{y}{2} = \frac{z}{3} = \mu$.
Then the coordinates of any point on the line $AD$ can be expressed as:
$x = 4 - 2\mu$
$y = 2\mu$
$z = 3\mu$

2. Substitute the parametric coordinates into the equation of plane $CBD$ to find the parameter at the intersection.

Since $D$ is a vertex of the tetrahedron, it must lie on the plane $CBD$. From the context given in part (ii), the equation of plane $CBD$ is $7x - 11y - 5z = -23$.
Substitute the coordinates of the line into the plane equation:
$7(4 - 2\mu) - 11(2\mu) - 5(3\mu) = -23$
$28 - 14\mu - 22\mu - 15\mu = -23$
$28 - 51\mu = -23$
$51\mu = 51 \implies \mu = 1$

3. Calculate the coordinates of point $D$.

Substitute $\mu = 1$ back into the parametric equations:
$x = 4 - 2(1) = 2$
$y = 2(1) = 2$
$z = 3(1) = 3$

The coordinates of $D$ are $(2, 2, 3)$.

1. Determine the coordinates of $A$, $B$, and $D$ using conditions given in the earlier parts of the question.

From part (i), $A$ lies on the $x$-axis ($y=z=0$) and on plane $ABD$ ($4x+y+2z=16$).
$4x = 16 \implies x = 4 \implies A(4, 0, 0)$.

From part (iii), $B$ lies on the $xy$-plane ($z=0$), plane $ABD$, and plane $CBD$ ($7x-11y-5z=-23$).
Solving $4x+y=16$ and $7x-11y=-23$ yields $x=3, y=4 \implies B(3, 4, 0)$.

From part (iv), $D$ lies on line $AD$ ($x=4-2\mu, y=2\mu, z=3\mu$) and plane $CBD$.
Substitute into plane $CBD$:
$7(4-2\mu) - 11(2\mu) - 5(3\mu) = -23 \implies 28 - 51\mu = -23 \implies \mu = 1 \implies D(2, 2, 3)$.

2. Find the vectors $\vec{AB}$ and $\vec{AC}$, and calculate the exact area of triangle $ABC$.

$\vec{AB} = \begin{pmatrix} 3 - 4 \\ 4 - 0 \\ 0 - 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ 0 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} -1 - 4 \\ 1 - 0 \\ 1 - 0 \end{pmatrix} = \begin{pmatrix} -5 \\ 1 \\ 1 \end{pmatrix}$

$\vec{AB} \times \vec{AC} = \begin{pmatrix} -1 \\ 4 \\ 0 \end{pmatrix} \times \begin{pmatrix} -5 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4(1) - 0(1) \\ 0(-5) - (-1)(1) \\ (-1)(1) - 4(-5) \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 19 \end{pmatrix}$

$\text{Area of triangle } ABC = \frac{1}{2}|\vec{AB} \times \vec{AC}|$
$= \frac{1}{2}\sqrt{4^2 + 1^2 + 19^2} = \frac{1}{2}\sqrt{16 + 1 + 361} = \frac{\sqrt{378}}{2}$

3. Find the perpendicular height from $D$ to the plane $ABC$.

Let the normal vector to plane $ABC$ be $\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 4 \\ 1 \\ 19 \end{pmatrix}$.
$\vec{AD} = \begin{pmatrix} 2 - 4 \\ 2 - 0 \\ 3 - 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ 3 \end{pmatrix}$

The perpendicular height $h$ is the length of the projection of $\vec{AD}$ onto $\mathbf{n}$:
$h = \frac{|\vec{AD} \cdot \mathbf{n}|}{|\mathbf{n}|} = \frac{\left| \begin{pmatrix} -2 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 1 \\ 19 \end{pmatrix} \right|}{\sqrt{378}}$
$= \frac{|-8 + 2 + 57|}{\sqrt{378}} = \frac{51}{\sqrt{378}}$

4. Calculate the volume of the tetrahedron.

$\text{Volume} = \frac{1}{3} \times \text{Area of base} \times h$
$= \frac{1}{3} \times \frac{\sqrt{378}}{2} \times \frac{51}{\sqrt{378}}$
$= \frac{51}{6} = \frac{17}{2}$