Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

Find how many sets of three points can be selected which can form the vertices of a triangle.

Verify that $\text{P}$ is the point of intersection of the two lines.

Write down the value of $\lambda$ corresponding to the point $\text{A}$.

Write down $\overrightarrow{\text{PA}}$ and $\overrightarrow{\text{PB}}$.

Let $\text{C}$ be the point on $l_1$ with coordinates (1, 0, 1) and $\text{D}$ be the point on $l_2$ with parameter $\mu =-2$.

Find the area of the quadrilateral $\text{CDBA}$.

appreciation that two points distinct from $\text{P}$ need to be chosen from each line   M1

${}^4{C}_2\times {}^3{C}_2$

=18    A1

[2 marks]

EITHER

consider cases for triangles including $\text{P}$ or triangles not including $\text{P}$      M1

$3\times 4+4\times {}^3{C}_2+3\times {}^4{C}_2$     (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

OR

consider total number of ways to select 3 points and subtract those with 3 points on the same line      M1

${}^8{C}_3-{}^5{C}_3-{}^4{C}_3$     (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42    A1

[4 marks]

METHOD 1

substitution of (4, 6, 4) into both equations       (M1)

$\lambda =3$ and $\mu =1$       A1A1

(4, 6, 4)       AG

METHOD 2

attempting to solve two of the three parametric equations      M1

$\lambda =3$ and $\mu =1$       A1

check both of the above give (4, 6, 4)       M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of "$\lambda$" or "$\mu$".

[3 marks]

$\lambda =2$      A1

[1 mark]

$\overrightarrow{\text{PA}}=\left(\begin{array}{c}-1\\ -2\\ -1\end{array}\right)$ ,  $\overrightarrow{\text{PB}}=\left(\begin{array}{c}-5\\ -6\\ -2\end{array}\right)$    A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

METHOD 1

area triangle $\text{ABP}=\frac{1}{2}\left|\overrightarrow{\text{PB}}\times \overrightarrow{\text{PA}}\right|$    M1

$\left(=\frac{1}{2}\left|\left(\begin{array}{c}-5\\ -6\\ -2\end{array}\right)\times \left(\begin{array}{c}-1\\ -2\\ -1\end{array}\right)\right|\right)=\frac{1}{2}\left|\left(\begin{array}{c}2\\ -3\\ 4\end{array}\right)\right|$    A1

$=\frac{\sqrt{29}}{2}$    A1

EITHER

$\overrightarrow{\text{PC}}=3\overrightarrow{\text{PA}}$ , $\overrightarrow{\text{PD}}=3\overrightarrow{\text{PB}}$       (M1)

area triangle $\text{PCD}=9\times$ area triangle $\text{ABP}$       (M1)A1

$=\frac{9\sqrt{29}}{2}$    A1

OR

$\text{D}$ has coordinates (−11, −12, −2)    A1

area triangle $\text{PCD}=\frac{1}{2}\left|\overrightarrow{\text{PD}}\times \overrightarrow{\text{PC}}\right|=\frac{1}{2}\left|\left(\begin{array}{c}-15\\ -18\\ -6\end{array}\right)\times \left(\begin{array}{c}-3\\ -6\\ -3\end{array}\right)\right|$    M1A1

Note: A1 is for the correct vectors in the correct formula.

$=\frac{9\sqrt{29}}{2}$    A1

THEN

area of $\text{CDBA}=\frac{9\sqrt{29}}{2}-\frac{\sqrt{29}}{2}$

$=4\sqrt{29}$    A1

METHOD 2

$\text{D}$ has coordinates (−11, −12, −2)    A1

area $=\frac{1}{2}\left|\overrightarrow{\text{CB}}\times \overrightarrow{\text{CA}}\right|+\frac{1}{2}\left|\overrightarrow{\text{BC}}\times \overrightarrow{\text{BD}}\right|$      M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

$\overrightarrow{\text{CB}}=\left(\begin{array}{c}-2\\ 0\\ 1\end{array}\right)$ , $\overrightarrow{\text{CA}}=\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)$    A1

$\overrightarrow{\text{CB}}\times \overrightarrow{\text{CA}}=\left(\begin{array}{c}-4\\ 6\\ -8\end{array}\right)$    A1

$\overrightarrow{\text{BC}}=\left(\begin{array}{c}2\\ 0\\ -1\end{array}\right)$ , $\overrightarrow{\text{BD}}=\left(\begin{array}{c}-10\\ -12\\ -4\end{array}\right)$    A1

$\overrightarrow{\text{BC}}\times \overrightarrow{\text{BD}}=\left(\begin{array}{c}-12\\ 18\\ -24\end{array}\right)$    A1

Note: Other vectors which might be used are $\overrightarrow{\text{DA}}=\left(\begin{array}{c}14\\ 16\\ 5\end{array}\right)$ , $\overrightarrow{\text{BA}}=\left(\begin{array}{c}4\\ 4\\ 1\end{array}\right)$, $\overrightarrow{\text{DC}}=\left(\begin{array}{c}12\\ 12\\ 3\end{array}\right)$.

Note: Previous A1A1A1A1 are all dependent on the first M1.

valid attempt to find a value of $\frac{1}{2}\left|a\times b\right|$      M1

Note: M1 independent of triangle chosen.

area $=\frac{1}{2}\times 2\times \sqrt{29}+\frac{1}{2}\times 6\times \sqrt{29}$

$=4\sqrt{29}$    A1

Note: accept $\frac{1}{2}\sqrt{116}+\frac{1}{2}\sqrt{1044}$ or equivalent.

[8 marks]