(a)

Method 1: Cross Product

The plane’s normal vector is the cross product of direction vectors \(\begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}\) and \(\begin{pmatrix} 6 \\ -3 \\ 2 \end{pmatrix}\):

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 2 \\ 6 & -3 & 2 \end{vmatrix} = \mathbf{i}(3 \cdot 2 – 2 \cdot (-3)) – \mathbf{j}(2 \cdot 2 – 2 \cdot 6) + \mathbf{k}(2 \cdot (-3) – 3 \cdot 6) \] (M1)(A1)

\[ = \begin{pmatrix} 12 \\ 8 \\ -24 \end{pmatrix} = 4 \begin{pmatrix} 3 \\ 2 \\ -6 \end{pmatrix} \] (A1)

Plane equation: \[ 3x + 2y – 6z = d \] (M1)

Using point (-2, 3, -2): \[ 3(-2) + 2(3) – 6(-2) = -6 + 6 + 12 = 12 \] (A1)

\[ 3x + 2y – 6z = 12 \] (A1)

Method 2: Elimination

Parametric equations:

\[ x = -2 + 2\lambda + 6\mu, \quad y = 3 + 3\lambda – 3\mu, \quad z = -2 + 2\lambda + 2\mu \] (M1)

Eliminate \(\mu\): \[ \mu = \frac{x + 2 – 2\lambda}{6} \]

Substitute into \(y\): \[ y = 3 + 3\lambda – 3 \cdot \frac{x + 2 – 2\lambda}{6} \implies x + 2y = 8\lambda + 4 \] (A1)

Substitute into \(z\): \[ 3z = -6 + 6\lambda + (x + 2 – 2\lambda) \implies x + 3z = 4\lambda – 4 \] (A1)

Eliminate \(\lambda\): \[ 3(x + 2y) – 2(x + 3z) = 3(8\lambda + 4) – 2(4\lambda – 4) \implies 3x + 2y – 6z = 12 \] (M1)(A1)(A1)

[6 marks]

(b)

Plane: \(3x + 2y – 6z = 12\).

x-axis (\(y = 0, z = 0\)): \[ 3x = 12 \implies x = 4 \]. A = (4, 0, 0) (A1)

y-axis (\(x = 0, z = 0\)): \[ 2y = 12 \implies y = 6 \]. B = (0, 6, 0) (A1)

z-axis (\(x = 0, y = 0\)): \[ -6z = 12 \implies z = -2 \]. C = (0, 0, -2) (A1)

[3 marks]

(c)

Volume of pyramid OABC: \(V = \frac{1}{6} |\mathbf{OA} \cdot (\mathbf{OB} \times \mathbf{OC})|\).

Vectors: \(\mathbf{OA} = \begin{pmatrix} 4 \\ 0 \\ 0 \end{pmatrix}\), \(\mathbf{OB} = \begin{pmatrix} 0 \\ 6 \\ 0 \end{pmatrix}\), \(\mathbf{OC} = \begin{pmatrix} 0 \\ 0 \\ -2 \end{pmatrix}\).

\[ \mathbf{OB} \times \mathbf{OC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 6 & 0 \\ 0 & 0 & -2 \end{vmatrix} = \begin{pmatrix} -12 \\ 0 \\ 0 \end{pmatrix} \] (M1)

\[ \mathbf{OA} \cdot \begin{pmatrix} -12 \\ 0 \\ 0 \end{pmatrix} = -48 \] (M1)

\[ V = \frac{1}{6} \cdot |-48| = 8 \] (A1)

[3 marks]

(d)

Angle between plane and x-axis is \(90^\circ – \phi\), where \(\phi\) is the angle between the normal \(\begin{pmatrix} 3 \\ 2 \\ -6 \end{pmatrix}\) and x-axis direction \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\):

\[ \cos \phi = \frac{3 \cdot 1 + 2 \cdot 0 + (-6) \cdot 0}{\sqrt{3^2 + 2^2 + (-6)^2} \cdot 1} = \frac{3}{7} \] (M1)(A1)

\[ \theta = 90^\circ – \cos^{-1}\left(\frac{3}{7}\right) \approx 25.4^\circ \text{ (or 0.443 radians)} \] (M1)(A1)

[4 marks]

(e)

Distance from origin to plane \(3x + 2y – 6z = 12\):

\[ d = \frac{|12|}{\sqrt{3^2 + 2^2 + (-6)^2}} = \frac{12}{7} \approx 1.71 \] (M1)(A1)

[2 marks]

(f)

Volume: \(V = \frac{1}{3} \cdot \text{Area of ABC} \cdot \text{Distance}\).

\[ 8 = \frac{1}{3} \cdot \text{Area} \cdot \frac{12}{7} \implies \text{Area} = 8 \cdot 3 \cdot \frac{7}{12} = 14 \] (M1)(A1)

[2 marks]