**(a)** | $\overrightarrow{OP} = 2\mathbf{i} + 5\mathbf{j} - \mathbf{k}$ or $\overrightarrow{OP} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$ | A1 | M1 For attempting one of $\overline{XB}$ or $\overline{BX}$ or $\overline{XA}$ or $\overline{AX}$. It must be correct for at least one of the components. E.g. $\overline{XB} = \overline{OB} - \overline{OX} = \begin{pmatrix} 3 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$ or $\overline{XA} = \overline{OA} - \overline{OX} = \begin{pmatrix} 0 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}$ either way round. |

**(b)** | Shows that $\overline{XB} = \overline{XA} = 3$ and states $XAPB$ is a rhombus. | A1 | M1 Attempts both $|\overline{XB}| = \sqrt{2^2 + 1^2 + (\pm2)^2}$ and $|\overline{XA}| = \sqrt{(\pm1)^2 + 2^2 + (\pm2)^2}$. If $\overline{XA}$ or $\overline{XB}$ has not been found in part a, it need to be calculated in part b. Alternatively attempts $\overline{XP} \cdot \overline{BA}$ or $XM^2, MB^2$ and $XB^2$ where $M$ is the mid point of $XP$. A1 Shows that $\overline{XB} = \overline{XA} = 3$ and states $XAPB$ is a rhombus. Requires both a reason and a conclusion. In the alternatives $\overline{XP} \cdot \overline{BA} = (\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j}) = -3 + 3 + 0 = 0$ so diagonals cross at $90°$ so $XAPB$ is a rhombus or $XM^2 + MB^2 = XB^2 = 6.5 + 2.5 = 9 \Rightarrow \angle XMB = 90° \Rightarrow$ rhombus |

**(c)** | $\sqrt{65}$ | A1 | M1 Attempts to find both $\overline{XP} = \overline{XB} + \overline{XA} = (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = (\mathbf{i} + 3\mathbf{j} - 4\mathbf{k})$ and $\overline{BA} = \overline{BX} + \overline{XA} = (-2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + (-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = (-3\mathbf{i} + \mathbf{j})$. You may see $\overline{XM} = \frac{1}{2}\overline{XB} + \frac{1}{2}\overline{BP} = \frac{1}{2}(2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) + \frac{1}{2}(-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})$ and $\overline{BM} = \frac{1}{2}\overline{BX} + \frac{1}{2}\overline{XA} = \frac{1}{2}(-2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + \frac{1}{2}(-\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) = \left(-\frac{3}{2}\mathbf{i} + \frac{1}{2}\mathbf{j}\right)$. M1 Attempts to find the area $XAPB$. E.g. Area $= \frac{1}{2}|\overline{XP}| \times |\overline{BA}| = \frac{1}{2} \times \sqrt{1^2 + 3^2 + (\pm4)^2} \times \sqrt{(\pm3)^2 + 1^2} = \frac{1}{2}\sqrt{26} \times \sqrt{10}$. Alternatively Alternatively the sum of the area of four right angled triangles. E.g. Area $= 4 \times \frac{1}{2} \times |\overline{XM}| \times |\overline{BM}| = 4 \times \frac{1}{2} \times \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2 + (\pm 2)^2} \times \sqrt{\left(\pm\frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$. Area $= 4 \times \frac{1}{2} \times \sqrt{\frac{13}{2}} \times \sqrt{\frac{5}{2}}$. A1 Correct answer only $\sqrt{65}$. Alternatively M1 Attempts to find the angle $\cos XBP$ or $\cos BPA$ using the formula $\overline{BX} \cdot \overline{BP} = \frac{1}{2}|\overline{BX}| \times |\overline{BP}| \cos XBP$. E.g. $\pm \begin{pmatrix} -2 \\ -1 \\ 2 \end{pmatrix} \cdot \pm \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \sqrt{(\pm2)^2 + (\pm1)^2 + (2)^2} \times \sqrt{(\pm1)^2 + 2^2 + (\pm2)^2} \cos XBP$, $2 - 2 - 4 = 3 \times 3 \times \cos XBP \Rightarrow \cos XBP = -\frac{4}{9}$. M1 Constructs a rigorous method leading to the area $XAPB$. E.g. Area $XAPB = 2 \times \frac{1}{2} \times$ Area of triangle $XBP = 2 \times \frac{1}{2} \times \sqrt{9} \times \sqrt{9} \sin XBP$, where $\sin XBP = \sqrt{1 - \cos^2 XBP} = \sqrt{1 - \left(-\frac{4}{9}\right)^2} = \frac{\sqrt{65}}{9}$. A1 Correct answer only $\sqrt{65}$. |

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