**(i)** $\mathbf{n} = l_1 \times l_2$ | B1 | For stating or implying in (i) or (ii) that n is perpendicular to $l_1$ and $l_2$

$\mathbf{n} = [2, -1, 1] \times [4, 3, 2]$ | M1* | For finding vector product of direction vectors

$\mathbf{n} = k[-1, 0, 2]$ | A1 | For correct vector (any $k$)

$[3, 4, -1] \cdot k[-1, 0, 2] = -5k$ | M1 | For substituting a point of $l_1$ into $\mathbf{r} \cdot \mathbf{n}$

$\mathbf{r} \cdot [-1, 0, 2] = -5$ | A1 5 | For obtaining correct $p$. AEF in this form

**(ii)** $[5, 1, 1] \cdot k[-1, 0, 2] = -3k$

$\mathbf{r} \cdot [-1, 0, 2] = -3$ | M1 | For using same $\mathbf{n}$ and substituting a point of $l_2$
| A1√ 2 | For obtaining correct $p$. AEF in this form f.t. on incorrect $\mathbf{n}$

**(iii)** $d = \frac{|-5+3|}{\sqrt{5}}$ OR $d = \frac{|[2, -3, 2] \cdot [-1, 0, 2]|}{\sqrt{5}}$ | M1 | For using a distance formula from their equations Allow omission of $| |$

OR $d$ from $(5, 1, 1)$ to $\Pi_1 = \frac{|5(-1) + 1(0) + 1(2) + 5|}{\sqrt{5}}$

OR $d$ from $(3, 4, -1)$ to $\Pi_2 = \frac{|3(-1) + 4(0) + (-1)(2) + 3|}{\sqrt{5}}$

OR $[3-t, 4--, 1+2t] \cdot [-1, 0, 2] = -5 \Rightarrow t = \frac{2}{5}$

OR $[5-t, 1, 1+2t] \cdot [-1, 0, 2] = -5 \Rightarrow t = -\frac{2}{5}$

$d = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} = 0.894427...$ | A1√ 2 | For correct distance AEF f.t. on incorrect $\mathbf{n}$

**(iv)** $d$ is the shortest OR perpendicular distance between $l_1$ and $l_2$ | B1 1 | For correct statement