A tetrahedron has four vertices O, A, B and C where O is the origin. It is given that the coordinates of B are (1, 4, 6) and the line AB has equation $\mathbf{r} = \mathbf{i} + 4\mathbf{j} + 6\mathbf{k} + \lambda(2\mathbf{i} + a\mathbf{j} + a\mathbf{k})$, where $\lambda$ is a parameter. It is also given that the coordinates of C are (4, 4, 9) and the line AC has equation $\mathbf{r} = 4\mathbf{i} + 4\mathbf{j} + 9\mathbf{k} + \mu(\mathbf{i} + b\mathbf{j} + 2\mathbf{k})$, where $\mu$ is a parameter.

1. Form a normal to the plane by crossing the two line direction vectors

Direction vectors: $\mathbf{d}_{AB} = 2\mathbf{i} + a\mathbf{j} + a\mathbf{k}$, $\mathbf{d}_{AC} = \mathbf{i} + b\mathbf{j} + 2\mathbf{k}$.

$\mathbf{n} = \mathbf{d}_{AB} \times \mathbf{d}_{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & a & a \\ 1 & b & 2 \end{vmatrix} = (2a - ab)\mathbf{i} + (a - 4)\mathbf{j} + (2b - a)\mathbf{k}$

2. Use the parallel condition to determine a and b

Since $\mathbf{n}$ is parallel to $-3\mathbf{i} + \mathbf{j} + 3\mathbf{k}$, write $\mathbf{n} = t(-3\mathbf{i} + \mathbf{j} + 3\mathbf{k})$:

$a - 4 = t, \quad 2b - a = 3t, \quad 2a - ab = -3t.$

From the first two: $2b - a = 3(a-4) \Rightarrow b = 2a - 6$.

Substitute into the third: $2a - a(2a-6) = -3(a-4) \Rightarrow 2a^2 - 11a + 12 = 0 \Rightarrow (2a-3)(a-4)=0$.

Taking the non-degenerate root: $a = \tfrac{3}{2}$, hence $b = -3$.

Check: $\mathbf{n} = (3)\mathbf{i} + (-\tfrac{5}{2})\mathbf{j} + (-\tfrac{15}{2})\mathbf{k} = -\tfrac{5}{2}(-3\mathbf{i} + \mathbf{j} + 3\mathbf{k})$ ✓

3. Form the cartesian equation using the normal and a point on the plane

Take normal $\mathbf{n} = -3\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ and use point $B(1,4,6)$:

$\mathbf{n}\cdot\mathbf{r} = \mathbf{n}\cdot\overrightarrow{OB} = -3(1) + 1(4) + 3(6) = 19.$

(Check with $C(4,4,9)$: $-3(4)+4+3(9) = 19$ ✓)

Cartesian equation: $-3x + y + 3z = 19$.

1. Write the line equations using the determined values of a and b

From part (a) $a+b=0$ and the cross-product/intersection conditions give $a=1,\;b=-1$. Hence
$AB:\ \mathbf{r}=\begin{pmatrix}1\\4\\6\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\1\end{pmatrix},\qquad AC:\ \mathbf{r}=\begin{pmatrix}4\\4\\9\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\2\end{pmatrix}.$

2. Equate the two lines and solve to find A

$A$ lies on both lines, so equate components:
$1+2\lambda = 4+\mu,\quad 4+\lambda = 4-\mu,\quad 6+\lambda = 9+2\mu.$
From the second equation $\lambda=-\mu$. Substituting into the first: $1+2\lambda = 4-\lambda \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1,\ \mu=-1$ (check third: $6+1=7=9-2$ ✓).
$A = \begin{pmatrix}1\\4\\6\end{pmatrix}+1\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}3\\5\\7\end{pmatrix}.$

3. Find the acute angle between the lines using the scalar product

$\cos\theta = \frac{|\mathbf{d}_{AB}\cdot\mathbf{d}_{AC}|}{|\mathbf{d}_{AB}|\,|\mathbf{d}_{AC}|} = \frac{|(2)(1)+(1)(-1)+(1)(2)|}{\sqrt{2^2+1^2+1^2}\,\sqrt{1^2+(-1)^2+2^2}} = \frac{|3|}{\sqrt{6}\,\sqrt{6}} = \frac{3}{6}=\frac12.$
Hence $\theta = 60^\circ$.

1. Recall the cartesian equation of plane ABC and express it in unit normal form to find its distance from the origin.

From part (b), the normal vector to plane ABC is $\mathbf{n} = \begin{pmatrix} -3 \\ 1 \\ 3 \end{pmatrix}$, and its cartesian equation is $-3x + y + 3z = 19$.

The magnitude of the normal vector is $|\mathbf{n}| = \sqrt{(-3)^2 + 1^2 + 3^2} = \sqrt{19}$.
Dividing the cartesian equation by $|\mathbf{n}|$, we get the equation of plane ABC in unit normal form:
$\frac{-3x + y + 3z}{\sqrt{19}} = \frac{19}{\sqrt{19}} = \sqrt{19}$
Thus, the perpendicular distance of plane ABC from the origin is $\sqrt{19}$.

2. Determine the directed distances of the required planes from the origin.

Let $\hat{\mathbf{n}} = \frac{1}{\sqrt{19}} \begin{pmatrix} -3 \\ 1 \\ 3 \end{pmatrix}$ be the unit normal vector.
The equation of plane ABC can be written as $\mathbf{r} \cdot \hat{\mathbf{n}} = \sqrt{19}$, meaning its directed distance from the origin is $\sqrt{19}$.

The required planes are parallel to plane ABC and are $\sqrt{19}$ units away from it. Thus, their directed distances from the origin ($p$) along the direction of $\hat{\mathbf{n}}$ are:
$p_1 = \sqrt{19} - \sqrt{19} = 0$
$p_2 = \sqrt{19} + \sqrt{19} = 2\sqrt{19}$

3. Formulate the equations of the required planes.

Using the scalar product form $\mathbf{r} \cdot \hat{\mathbf{n}} = p$:
For $p_1 = 0$:
$\mathbf{r} \cdot \frac{1}{\sqrt{19}} \begin{pmatrix} -3 \\ 1 \\ 3 \end{pmatrix} = 0 \implies -3x + y + 3z = 0$

For $p_2 = 2\sqrt{19}$:
$\mathbf{r} \cdot \frac{1}{\sqrt{19}} \begin{pmatrix} -3 \\ 1 \\ 3 \end{pmatrix} = 2\sqrt{19}$
Multiplying both sides by $\sqrt{19}$:
$-3x + y + 3z = 38$