平面 $\pi$ 通过坐标为 $(-1,0,1)$，$(2,1,-1)$ 和 $(1,-3,2)$ 的点。直线 $l$ 通过点 $P(16,20,16)$ 且平行于向量 $\begin{pmatrix}1 \\ 2 \\ 1\end{pmatrix}$。

1. Find two non-parallel direction vectors lying on the plane $\pi$.

Let $A(-1, 0, 1)$, $B(2, 1, -1)$ and $C(1, -3, 2)$ be points on the plane.
$\vec{AB} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} - \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$

2. Calculate the normal vector to the plane by taking the cross product of the two direction vectors.

Let $\mathbf{n}$ be a normal vector to $\pi$.
$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (1)(1) - (-2)(-3) \\ (-2)(2) - (3)(1) \\ (3)(-3) - (1)(2) \end{pmatrix} = \begin{pmatrix} 1 - 6 \\ -4 - 3 \\ -9 - 2 \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ -11 \end{pmatrix}$
We can take a simpler parallel normal vector $\mathbf{n}_1 = \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$.

3. Find the cartesian equation of the plane using the scalar product form $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.

The equation of the plane is given by $\mathbf{r} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix} = \vec{OA} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$
$5x + 7y + 11z = (-1)(5) + (0)(7) + (1)(11)$
$5x + 7y + 11z = 6$

1. Concisely obtain the cartesian equation of $\pi$ (required from part i).

From the given points $A(-1,0,1)$, $B(2,1,-1)$, and $C(1,-3,2)$, we find the direction vectors:
$\vec{AB} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}, \quad \vec{AC} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$
The normal to the plane is:
$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} -5 \\ -7 \\ -11 \end{pmatrix} = -\begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$
The cartesian equation of $\pi$ is $\mathbf{r} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$, which simplifies to:
$5x + 7y + 11z = 6$

2. Formulate the parametric equations of the line $l$.

The vector equation of line $l$ is:
$\mathbf{r} = \begin{pmatrix} 16 \\ 20 \\ 16 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$
So the parametric equations are $x = 16 + \lambda$, $y = 20 + 2\lambda$, and $z = 16 + \lambda$.

3. Substitute the line equations into the plane equation and solve for $\lambda$.

$5(16 + \lambda) + 7(20 + 2\lambda) + 11(16 + \lambda) = 6$
$80 + 5\lambda + 140 + 14\lambda + 176 + 11\lambda = 6$
$396 + 30\lambda = 6$
$30\lambda = -390 \implies \lambda = -13$

4. Find the point of intersection.

Substituting $\lambda = -13$ back into the equation of $l$:
$\mathbf{r} = \begin{pmatrix} 16 \\ 20 \\ 16 \end{pmatrix} - 13 \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 3 \end{pmatrix}$
The coordinates of the point of intersection are $(3, -6, 3)$.

1. Find the normal vector to the plane $\pi$.

Two vectors on the plane $\pi$ are $\vec{AB} = \begin{pmatrix} 2 - (-1) \\ 1 - 0 \\ -1 - 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}$ and $\vec{AC} = \begin{pmatrix} 1 - (-1) \\ -3 - 0 \\ 2 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$.
The normal vector $\mathbf{n}$ is given by their cross product:
$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(1) - (-2)(-3) \\ -2(2) - 3(1) \\ 3(-3) - 1(2) \end{pmatrix} = \begin{pmatrix} -5 \\ -7 \\ -11 \end{pmatrix}$
We can take the normal vector as $\mathbf{n} = \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix}$.

2. Calculate the acute angle between the line $l$ and the plane $\pi$.

The line $l$ is parallel to its direction vector $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. Let $\theta$ be the acute angle between the line and the plane. We use the relation $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$:
$\sin \theta = \frac{\left| \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ 7 \\ 11 \end{pmatrix} \right|}{\sqrt{1^2+2^2+1^2}\sqrt{5^2+7^2+11^2}}$
$\sin \theta = \frac{5(1) + 7(2) + 11(1)}{\sqrt{6}\sqrt{195}} = \frac{30}{\sqrt{1170}} = \frac{10}{\sqrt{130}}$
$\theta = \sin^{-1}\left(\frac{10}{\sqrt{130}}\right) \approx 61.289^\circ$

1. Find the Cartesian equation of $\pi$ and the intersection point $I$ of $l$ and $\pi$ (intermediate results normally obtained in parts i and ii).

Normal to $\pi$: $\mathbf{n} = \begin{pmatrix}2 - (-1) \\ 1 - 0 \\ -1 - 1\end{pmatrix} \times \begin{pmatrix}1 - (-1) \\ -3 - 0 \\ 2 - 1\end{pmatrix} = \begin{pmatrix}3 \\ 1 \\ -2\end{pmatrix} \times \begin{pmatrix}2 \\ -3 \\ 1\end{pmatrix} = \begin{pmatrix}5 \\ 7 \\ 11\end{pmatrix}$.
Equation of $\pi$: $\begin{pmatrix}x \\ y \\ z\end{pmatrix} \cdot \begin{pmatrix}5 \\ 7 \\ 11\end{pmatrix} = \begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix} \cdot \begin{pmatrix}5 \\ 7 \\ 11\end{pmatrix} \implies 5x + 7y + 11z = 6$.
Substitute $l$: $\mathbf{r} = \begin{pmatrix}16\\20\\16\end{pmatrix} + \lambda\begin{pmatrix}1\\2\\1\end{pmatrix}$ into $\pi$:
$5(16+\lambda) + 7(20+2\lambda) + 11(16+\lambda) = 6 \implies 30\lambda + 396 = 6 \implies \lambda = -13$.
Intersection point $I = \begin{pmatrix}16 - 13 \\ 20 - 26 \\ 16 - 13\end{pmatrix} = \begin{pmatrix}3 \\ -6 \\ 3\end{pmatrix}$.

2. Find the position vector of the foot of the perpendicular, $F$, from point $P$ on $l$ to the plane $\pi$.

Let $P(16, 20, 16)$ be a point on $l$. The line passing through $P$ and perpendicular to $\pi$ has direction vector $\mathbf{n}$.
Equation of $PF$: $\mathbf{r} = \begin{pmatrix}16\\20\\16\end{pmatrix} + \mu\begin{pmatrix}5\\7\\11\end{pmatrix}$.
Substitute into $\pi$ to find the intersection $F$:
$5(16+5\mu) + 7(20+7\mu) + 11(16+11\mu) = 6$
$396 + 195\mu = 6 \implies 195\mu = -390 \implies \mu = -2$.
$\vec{OF} = \begin{pmatrix}16\\20\\16\end{pmatrix} - 2\begin{pmatrix}5\\7\\11\end{pmatrix} = \begin{pmatrix}6\\6\\-6\end{pmatrix}$.

3. Find the position vector of the reflected point $P'$ in the plane $\pi$.

Since $F$ is the midpoint of the line segment joining $P$ to its reflection $P'$, we have:
$\vec{OF} = \frac{\vec{OP} + \vec{OP'}}{2} \implies \vec{OP'} = 2\vec{OF} - \vec{OP}$.
$\vec{OP'} = 2\begin{pmatrix}6\\6\\-6\end{pmatrix} - \begin{pmatrix}16\\20\\16\end{pmatrix} = \begin{pmatrix}12\\12\\-12\end{pmatrix} - \begin{pmatrix}16\\20\\16\end{pmatrix} = \begin{pmatrix}-4\\-8\\-28\end{pmatrix}$.

4. Find the direction vector of the reflected line and express its vector equation.

The reflected line passes through the point of intersection $I(3, -6, 3)$ and the reflected point $P'(-4, -8, -28)$.
Direction vector $\mathbf{d'} = \vec{IP'} = \begin{pmatrix}-4\\-8\\-28\end{pmatrix} - \begin{pmatrix}3\\-6\\3\end{pmatrix} = \begin{pmatrix}-7\\-2\\-31\end{pmatrix}$.
Taking a parallel direction vector $\begin{pmatrix}7\\2\\31\end{pmatrix}$, the vector equation of the line of reflection is:
$\mathbf{r} = \begin{pmatrix}3\\-6\\3\end{pmatrix} + t\begin{pmatrix}7\\2\\31\end{pmatrix}$, where $t \in \mathbb{R}$.