METHOD 1

\(\overrightarrow{{\text{AB}}} = \boldsymbol{b} – \boldsymbol{a} = \begin{pmatrix}1\\2\\3\end{pmatrix} – \begin{pmatrix}1\\1\\2\end{pmatrix} = \begin{pmatrix}0\\1\\1\end{pmatrix}\) (A1)

\(\overrightarrow{{\text{AC}}} = \boldsymbol{c} – \boldsymbol{a} = \begin{pmatrix}3\\0\\1\end{pmatrix} – \begin{pmatrix}1\\1\\2\end{pmatrix} = \begin{pmatrix}2\\-1\\-1\end{pmatrix}\) (A1)

\(\overrightarrow{{\text{AB}}} \times \overrightarrow{{\text{AC}}} = \begin{vmatrix}\boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k}\\0&1&1\\2&-1&-1\end{vmatrix}\) (M1)

= i(-1 + 1) – j(0 – 2) + k(0 – 2) (A1)

= 2j – 2k (A1)

Area of triangle ABC = \(\frac{1}{2}|2\mathbf{j} – 2\mathbf{k}| = \frac{1}{2}\sqrt{8} (= \sqrt{2})\) sq. units (M1)(A1)

METHOD 2

\(|\text{AB}| = \sqrt{2}\), \(|\text{BC}| = \sqrt{12}\), \(|\text{AC}| = \sqrt{6}\) (A1)(A1)(A1)

Using cosine rule, e.g. on \(\hat{C}\) (M1)

\(\cos C = \frac{6 + 12 – 2}{2\sqrt{72}} = \frac{2\sqrt{2}}{3}\) (A1)

\(\therefore \text{Area } \Delta \text{ABC} = \frac{1}{2}ab\sin C\) (M1)

\(= \frac{1}{2}\sqrt{12}\sqrt{6}\sin\left(\arccos\frac{2\sqrt{2}}{3}\right)\)

\(= 3\sqrt{2}\sin\left(\arccos\frac{2\sqrt{2}}{3}\right) (= \sqrt{2})\) (A1)

[7 marks]

\(\text{AB} = \sqrt{2}\) (A1)

\(\sqrt{2} = \frac{1}{2}\text{AB} \times h = \frac{1}{2}\sqrt{2} \times h\), where \(h\) equals the shortest distance (M1)

\(\Rightarrow h = 2\) (A1)

[3 marks]

METHOD 1

\(\pi\) has form \(r \cdot \begin{pmatrix}0\\2\\-2\end{pmatrix} = d\) (M1)

Since (1, 1, 2) is on the plane:

\(d = \begin{pmatrix}1\\1\\2\end{pmatrix} \cdot \begin{pmatrix}0\\2\\-2\end{pmatrix} = 2 – 4 = -2\) (M1)(A1)

Hence \(r \cdot \begin{pmatrix}0\\2\\-2\end{pmatrix} = -2\)

\(2y – 2z = -2\) (or \(y – z = -1\)) (A1)

METHOD 2

\(r = \begin{pmatrix}1\\1\\2\end{pmatrix} + \lambda \begin{pmatrix}0\\1\\1\end{pmatrix} + \mu \begin{pmatrix}2\\-1\\-1\end{pmatrix}\) (M1)

\(x = 1 + 2\mu\) (i)

\(y = 1 + \lambda – \mu\) (ii)

\(z = 2 + \lambda – \mu\) (iii) (A1)

From (i) \(\mu = \frac{x – 1}{2}\)

Substitute in (ii) \(y = 1 + \lambda – \left(\frac{x – 1}{2}\right)\)

\(\Rightarrow \lambda = y – 1 + \left(\frac{x – 1}{2}\right)\)

Substitute \(\lambda\) and \(\mu\) in (iii) (M1)

\(\Rightarrow z = 2 + y – 1 + \left(\frac{x – 1}{2}\right) – \left(\frac{x – 1}{2}\right)\)

\(\Rightarrow y – z = -1\) (A1)

[4 marks]

The equation of OD is:

\(r = \lambda \begin{pmatrix}0\\2\\-2\end{pmatrix}\), or \(r = \lambda \begin{pmatrix}0\\1\\-1\end{pmatrix}\) (M1)

This meets \(\pi\) where:

\(2\lambda + 2\lambda = -1\) (M1)

\(\lambda = -\frac{1}{4}\) (A1)

Coordinates of D are \(\left(0, -\frac{1}{2}, \frac{1}{2}\right)\) (A1)

[4 marks]

\(|\overrightarrow{\text{OD}}| = \sqrt{0 + \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{1}{\sqrt{2}}\) (M1)(A1)

[2 marks]