The line $l_1$ has the vector equation $\mathbf{r}=\begin{pmatrix}a\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}0\\1\\b\end{pmatrix},\ \lambda\in\mathbb{R},$ where $a$ and $b$ are real constants. The plane $p_1$ has cartesian equation $x+2y+4z=4$. The line $l_2$ has cartesian equation $\dfrac{1-y}{2}=z-3,\ x=5$. The line $l_1$ intersects the plane $p_1$ at the point with coordinates $(a,1,0)$.

1. Convert the cartesian equation of $l_2$ into vector form

Let $\dfrac{1-y}{2}=z-3=\mu$. Then $y=1-2\mu$, $z=3+\mu$, and $x=5$.

So a point on $l_2$ is $(5,\,1-2\mu,\,3+\mu)$, giving
$l_2:\ \mathbf{r}=\begin{pmatrix}5\\1\\3\end{pmatrix}+\mu\begin{pmatrix}0\\-2\\1\end{pmatrix},\ \mu\in\mathbb{R}.$
The direction vector of $l_2$ is $\begin{pmatrix}0\\-2\\1\end{pmatrix}$.

2. Find $a$ using the intersection point on the plane

$l_1$ meets $p_1$ at $(a,1,0)$, so this point satisfies $x+2y+4z=4$:
$a+2(1)+4(0)=4\ \Rightarrow\ a+2=4\ \Rightarrow\ a=2.$

3. Find $b$ using the perpendicular relationship between $l_1$ and $l_2$

Since $l_1$ and $l_2$ are perpendicular, their direction vectors have zero scalar product:
$\begin{pmatrix}0\\1\\b\end{pmatrix}\cdot\begin{pmatrix}0\\-2\\1\end{pmatrix}=0\ \Rightarrow\ 0-2+b=0\ \Rightarrow\ b=2.$
Hence $a=b=2$.

1. Substitute the coordinates of point $A$ and the values $a=b=2$ into the vector equation of $l_1$

$\begin{pmatrix}2\\2\\2\end{pmatrix} = \begin{pmatrix}2\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}0\\1\\2\end{pmatrix}$

2. Solve for $\lambda$ and verify consistency across all components

From the $y$-component:
$2 = 1 + \lambda \implies \lambda = 1$

Check $x$-component:
$2 = 2 + (1)(0)$ (consistent)

Check $z$-component:
$2 = 0 + (1)(2)$ (consistent)

Since there is a unique value of $\lambda = 1$ that satisfies all three equations, the point $A$ lies on $l_1$.

1. Set up the vector equation of the line passing through $A$ and perpendicular to $p_1$.

Let the line perpendicular to $p_1$ passing through $A(2,2,2)$ be $l_A$.
The normal vector to $p_1$ is $\mathbf{n} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$.
The equation of $l_A$ is:
$\mathbf{r} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}, \quad \mu \in \mathbb{R}$

2. Substitute the line equation into the plane equation to find the value of the parameter $\mu$ at the point of intersection.

Since the foot of the perpendicular $F$ lies on both $l_A$ and $p_1$, substitute the line's coordinates into the cartesian equation of the plane $x + 2y + 4z = 4$:
$(2 + \mu) + 2(2 + 2\mu) + 4(2 + 4\mu) = 4$
$2 + \mu + 4 + 4\mu + 8 + 16\mu = 4$
$14 + 21\mu = 4$
$21\mu = -10 \implies \mu = -\frac{10}{21}$

3. Calculate the coordinates of $F$.

Substitute $\mu = -\frac{10}{21}$ back into the equation of $l_A$ to find the position vector of $F$:
$\vec{OF} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} - \frac{10}{21} \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} \frac{32}{21} \\[6pt] \frac{22}{21} \\[6pt] \frac{2}{21} \end{pmatrix}$
The coordinates of $F$ are $\left( \frac{32}{21}, \frac{22}{21}, \frac{2}{21} \right)$.

4. Use $F$ to find the shortest distance from $A$ to $p_1$.

The shortest distance is the magnitude of the vector $\vec{AF}$.
$\vec{AF} = \vec{OF} - \vec{OA} = -\frac{10}{21} \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$
$|\vec{AF}| = \left|-\frac{10}{21}\right| \sqrt{1^2 + 2^2 + 4^2}$
$|\vec{AF}| = \frac{10}{21} \sqrt{21} = \frac{10\sqrt{21}}{21} \text{ units}$

1. Identify two non-parallel direction vectors that lie on the plane $p_2$.

Since $p_2$ contains the line $l_1$, its first direction vector is the direction vector of $l_1$, $\mathbf{d} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$.

Also, $p_2$ contains the point $A(2,2,2)$ (which lies on $l_1$) and the point $F$. Thus, the vector $\vec{AF}$ lies on $p_2$.
Because $F$ is the foot of the perpendicular from $A$ to $p_1$, $\vec{AF}$ is parallel to the normal of $p_1$, which is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$. This gives the second direction vector for $p_2$. (Note: We do not need the exact coordinates of $F$ to establish this direction).",

2. Find the normal vector to the plane $p_2$.

The normal vector $\mathbf{n}_2$ to $p_2$ is perpendicular to both direction vectors. We find it using the cross product:
$\mathbf{n}_2 = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} (1)(4) - (2)(2) \\ (2)(1) - (0)(4) \\ (0)(2) - (1)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix}$

3. Determine the scalar constant $d$ and state the vector equation of the plane $p_2$.

Using a point on $p_2$, such as $(2, 1, 0)$ from the line $l_1$:
$\mathbf{r} \cdot \mathbf{n}_2 = \mathbf{a} \cdot \mathbf{n}_2$
$d = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 0 + 2 + 0 = 2$

The vector equation of the plane $p_2$ is:
$\mathbf{r} \cdot \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} = 2$