平面 $\pi_1$ 的方程为 $\mathbf{r}\cdot\begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix}=11.$ 平面 $\pi_2$ 经过点 $A(1,-1,2),\; B(2,3,4)$ 和 $C(3,4,3).$

1. Find two direction vectors parallel to the plane $\pi_2$.

$\vec{AB} = \begin{pmatrix}2 \\ 3 \\ 4\end{pmatrix} - \begin{pmatrix}1 \\ -1 \\ 2\end{pmatrix} = \begin{pmatrix}1 \\ 4 \\ 2\end{pmatrix}$
$\vec{AC} = \begin{pmatrix}3 \\ 4 \\ 3\end{pmatrix} - \begin{pmatrix}1 \\ -1 \\ 2\end{pmatrix} = \begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix}$

2. Find a normal vector to the plane by calculating the cross product of the two direction vectors.

$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix}1 \\ 4 \\ 2\end{pmatrix} \times \begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix} = \begin{pmatrix} (4)(1) - (2)(5) \\ (2)(2) - (1)(1) \\ (1)(5) - (4)(2) \end{pmatrix} = \begin{pmatrix}-6 \\ 3 \\ -3\end{pmatrix}$
Dividing by $-3$ to use a simpler parallel normal vector, we choose $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

3. Formulate the cartesian equation of the plane using the normal vector and one of the given points.

The equation of the plane takes the form $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$. Using point $A$:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = (1)(2) + (-1)(-1) + (2)(1) = 5$
Expanding $\mathbf{r}$ as $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$, the cartesian equation is:
$2x - y + z = 5$

1. Determine the cartesian equation of $\pi_2$ (using results derived from part (i)).

From part (i), we use the given coordinates of points $A$, $B$, and $C$ to find the normal vector to $\pi_2$:
$\vec{AB} = \begin{pmatrix} 2-1 \\ 3-(-1) \\ 4-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}, \quad \vec{AC} = \begin{pmatrix} 3-1 \\ 4-(-1) \\ 3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix}$
$\mathbf{n}_2 = \vec{AB} \times \vec{AC} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 5 \\ 1 \end{pmatrix} = \begin{pmatrix} -6 \\ 3 \\ -3 \end{pmatrix} = -3\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
Taking $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ as the normal, the equation of $\pi_2$ is:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = 2(1) - 1(-1) + 1(2) = 5$
Hence, the cartesian equation of $\pi_2$ is $2x - y + z = 5$.

2. Solve the cartesian equations of $\pi_1$ and $\pi_2$ simultaneously to find a relationship between the coordinates on the line of intersection.

The cartesian equation of $\pi_1$ is directly obtained from its vector equation as $2x + 5y + z = 11$.
To find the line of intersection, we solve the system of equations:
$\begin{cases} 2x + 5y + z = 11 & \dots (1) \\ 2x - y + z = 5 & \dots (2) \end{cases}$
Subtracting (2) from (1) eliminates $x$ and $z$:
$6y = 6 \implies y = 1$
Substitute $y = 1$ into (2):
$2x - 1 + z = 5 \implies z = 6 - 2x$

3. Express the solution as a vector equation of the line.

Let $x = \lambda$, where $\lambda \in \mathbb{R}$. Then $z = 6 - 2\lambda$.
The position vector of a general point on the intersection line $l$ is:
$\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \lambda \\ 1 \\ 6 - 2\lambda \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

1. Express the displacement vector $\vec{AD}$ in terms of the line parameter $\lambda$.

Assuming from part (ii) the standard vector equation of the intersection line $l$ is $\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 6 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ (found by setting $x=0$), the position vector of a general point $D$ on $l$ is $\vec{OD} = \begin{pmatrix} \lambda \\ 1 \\ 6 - 2\lambda \end{pmatrix}$.

Given $A(1, -1, 2)$, the displacement vector $\vec{AD}$ is:
$\vec{AD} = \vec{OD} - \vec{OA} = \begin{pmatrix} \lambda \\ 1 \\ 6 - 2\lambda \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} \lambda - 1 \\ 2 \\ 4 - 2\lambda \end{pmatrix}$

2. Find the expression for the square of the distance $AD$.

$AD^2 = |\vec{AD}|^2 = (\lambda - 1)^2 + 2^2 + (4 - 2\lambda)^2$
$= (\lambda^2 - 2\lambda + 1) + 4 + (16 - 16\lambda + 4\lambda^2)$
$= 5\lambda^2 - 18\lambda + 21$

3. Find the value of $\lambda$ that gives the minimum distance.

To find the point nearest to $A$, we must minimise $AD^2$. Differentiating with respect to $\lambda$:
$\frac{\mathrm{d}(AD^2)}{\mathrm{d}\lambda} = 10\lambda - 18$
Set the derivative to $0$ for the stationary point (which gives minimum distance):
$10\lambda - 18 = 0 \implies \lambda = 1.8$

4. Determine the coordinates of the nearest point.

Substitute $\lambda = 1.8$ back into the position vector $\vec{OD}$:
$\vec{OD} = \begin{pmatrix} 1.8 \\ 1 \\ 6 - 2(1.8) \end{pmatrix} = \begin{pmatrix} 1.8 \\ 1 \\ 2.4 \end{pmatrix}$
Hence, the coordinates of the nearest point on $l$ to $A$ are $(1.8, 1, 2.4)$.

1. Formulate an expression for the distance from point $E$ to the plane $\pi_1$.

Since $\pi_3$ is parallel to $\pi_1$, the distance between the two planes is equal to the perpendicular distance from any point on $\pi_3$ (such as point $E$) to $\pi_1$.

The distance from point $E(1, k, 3)$ to the plane $\pi_1: \mathbf{r}\cdot\begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix}=11$ is given by:
$\text{Distance} = \frac{\left| \begin{pmatrix}1 \\ k \\ 3\end{pmatrix} \cdot \begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix} - 11 \right|}{\left| \begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix} \right|}$
$= \frac{|2(1) + 5(k) + 1(3) - 11|}{\sqrt{2^2 + 5^2 + 1^2}}$
$= \frac{|5k - 6|}{\sqrt{4 + 25 + 1}}$
$= \frac{|5k - 6|}{\sqrt{30}}$

2. Equate the distance expression to the given value and solve for $k$.

We are given that the distance is $\frac{16}{\sqrt{30}}$. Therefore:
$\frac{|5k - 6|}{\sqrt{30}} = \frac{16}{\sqrt{30}}$
$|5k - 6| = 16$

Solving the absolute value equation:
$5k - 6 = 16 \quad \text{or} \quad 5k - 6 = -16$
$5k = 22 \quad \text{or} \quad 5k = -10$
$k = \frac{22}{5} \quad \text{or} \quad k = -2$