The line l passes through the points A and B with coordinates (1, 2, 4) and (-2, 3, 1) respectively. The plane p has equation $3x - y + 2z = 17$. Find

1. Find the direction vector of the line $l$.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}$

2. Formulate the vector equation of the line $l$.

The equation of the line $l$ is given by:
$\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

3. Substitute the parametric equations of the line into the Cartesian equation of the plane $p$.

The Cartesian equation of the plane is $3x - y + 2z = 17$.
Substituting the parametric coordinates $x = 1 - 3\lambda$, $y = 2 + \lambda$, and $z = 4 - 3\lambda$:
$3(1 - 3\lambda) - (2 + \lambda) + 2(4 - 3\lambda) = 17$

4. Solve the linear equation for the parameter $\lambda$.

$3 - 9\lambda - 2 - \lambda + 8 - 6\lambda = 17$
$9 - 16\lambda = 17$
$-16\lambda = 8 \implies \lambda = -\frac{1}{2}$

5. Determine the coordinates of the point of intersection.

Substitute $\lambda = -\frac{1}{2}$ back into the parametric equations:
$x = 1 - 3\left(-\frac{1}{2}\right) = \frac{5}{2}$
$y = 2 + \left(-\frac{1}{2}\right) = \frac{3}{2}$
$z = 4 - 3\left(-\frac{1}{2}\right) = \frac{11}{2}$
The coordinates of the intersection point are $\left(\frac{5}{2}, \frac{3}{2}, \frac{11}{2}\right)$.

1. Find the direction vector of line $l$ and the normal vector of plane $p$.

The direction vector of line $l$ is given by $\vec{AB}$:
$\mathbf{d} = \vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}$
The normal vector to the plane $p$, read from its equation $3x - y + 2z = 17$, is:
$\mathbf{n} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$

2. Calculate the scalar product and the magnitudes of the vectors.

Let $\theta$ be the acute angle between line $l$ and plane $p$. The relationship using the scalar product is $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.

Calculate the scalar product:
$\mathbf{d} \cdot \mathbf{n} = \begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = (-3)(3) + (1)(-1) + (-3)(2) = -9 - 1 - 6 = -16$

Calculate the magnitudes:
$|\mathbf{d}| = \sqrt{(-3)^2 + 1^2 + (-3)^2} = \sqrt{9 + 1 + 9} = \sqrt{19}$
$|\mathbf{n}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

3. Solve for the acute angle $\theta$.

Substitute the values into the formula:
$\sin \theta = \frac{|-16|}{\sqrt{19}\sqrt{14}} = \frac{16}{\sqrt{266}}$
$\theta = \sin^{-1}\left(\frac{16}{\sqrt{266}}\right) \approx 78.813^\circ$
To 1 decimal place, the acute angle is $78.8^\circ$.

1. Identify the position vector of point A and the normal vector of plane p.

For point $A(1, 2, 4)$, its position vector is $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$.
The Cartesian equation of plane $p$ is $3x - y + 2z = 17$, which can be written as $\mathbf{r} \cdot \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = 17$.
Thus, its normal vector is $\mathbf{n} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ and $d = 17$.

2. Substitute the vectors into the formula for the perpendicular distance from a point to a plane.

The perpendicular distance from a point to a plane is given by the formula $\dfrac{|\mathbf{a} \cdot \mathbf{n} - d|}{|\mathbf{n}|}$:
$\text{Distance} = \frac{\left| \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} - 17 \right|}{\sqrt{3^2 + (-1)^2 + 2^2}}$

3. Evaluate the dot product and magnitude to find the exact distance.

$\text{Distance} = \frac{| 3(1) + (-1)(2) + 2(4) - 17 |}{\sqrt{9 + 1 + 4}}$
$= \frac{| 3 - 2 + 8 - 17 |}{\sqrt{14}}$
$= \frac{| -8 |}{\sqrt{14}}$
$= \frac{8}{\sqrt{14}}$
$= \frac{4\sqrt{14}}{7}$