## (i)
$(1, 3, 5)$ and $(5, 2, 5) \Rightarrow \pm[4, -1, 0]$ in $II$ | M1 | For finding a vector in $II$
$\mathbf{n} = [2, -2, 3] \times [4, -1, 0] = k[1, 4, 2]$ | M1 | For finding vector product of direction vectors of $l$ and a line in $II$
| A1 | For correct $\mathbf{n}$
$\Rightarrow r.[1, 4, 2] = 23$ | A1 4 | For correct equation. Allow multiples

## (ii) METHOD 1
Perpendicular to $II$ through $(-7, -3, 0)$ meets $II$ | M1 | For using perpendicular from point on $l$ to $II$. Award mark for $k\mathbf{n}$ used
| | 
where $(-7 + k) + 4(-3 + 4k) + 2(2k) = 23$ | M1 | For substituting parametric line coords into $II$
$\Rightarrow k = 2 \Rightarrow d = 2\sqrt{1^2 + 4^2 + 2^2} = 2\sqrt{21} = 9.165$ | M1 | For normalising the $\mathbf{n}$ used in this part
| A1 4 | For correct distance AEF

## (ii) METHOD 2
$II$ is $x + 4y + 2z = 23$ | M1 | For attempt to use formula for perpendicular distance
$\Rightarrow d = \frac{|(-7) + 4(-3) + 2(0) - 23|}{\sqrt{1^2 + 4^2 + 2^2}} = 2\sqrt{21} = 9.165$ | M1 | For substituting a point on $l$ into plane equation
| M1 | For normalising the $\mathbf{n}$ used in this part
| A1 | For correct distance AEF

## (ii) METHOD 3
$\mathbf{m} = [1, 3, 5] - [-7, -3, 0] = (\pm)[8, 6, 5]$ | M1 | For finding a vector from $l$ to $II$
OR $= [5, 2, 5] - [-7, -3, 0] = (\pm)[12, 5, 5]$ | | 
$\Rightarrow d = \frac{\mathbf{m}.[1, 4, 2]}{\sqrt{1^2 + 4^2 + 2^2}} = \frac{42}{\sqrt{21}} = 2\sqrt{21} = 9.165$ | M1 | For finding $\mathbf{m} \cdot \mathbf{n}$
| M1 | For normalising the $\mathbf{n}$ used in this part
| A1 | For correct distance AEF

## (ii) METHOD 4
$[-7, -3, 0] + k[1, 4, 2] = [1, 3, 5] + s[2, -2, 3] + t[4, -1, 0]$ | M1 | For using perpendicular from point on $l$ to $II$. Award mark for $k\mathbf{n}$ used

$k - 2s - 4t = 8$ | | 
$4k + 2s + t = 6$ | M1 | For setting up and solving 3 equations
$2k - 3s = 5$ | | 

$\Rightarrow k = 2\left(s = -\frac{1}{3}, t = -\frac{4}{3}\right)$ | M1 | For normalising the $\mathbf{n}$ used in this part
$\Rightarrow d = 2\sqrt{1^2 + 4^2 + 2^2} = 2\sqrt{21} = 9.165$ | A1 | For correct distance AEF

## (ii) METHOD 5
$d_1 = \frac{23}{\sqrt{1^2 + 4^2 + 2^2}} = \frac{23}{\sqrt{21}}$ | M1 | For attempt to find distance from O to $II$ OR from O to parallel plane containing $l$
$d_2 = \frac{[-7, -3, 0] \cdot [1, 4, 2]}{\sqrt{1^2 + 4^2 + 2^2}} = \frac{-19}{\sqrt{21}}$ | M1 | For normalising the $\mathbf{n}$ used in this part
$\Rightarrow d_1 - d_2 = d = \frac{23 - (-19)}{\sqrt{21}} = 2\sqrt{21} = 9.165$ | M1 | For finding $d_1 - d_2$
| A1 | For correct distance AEF

## (iii)
$(-7, -3, 0) + k(1, 4, 2)$ | M1 | State or imply coordinates of a point on the reflected line
Use $k = 4$ | M1 | State or imply 2 × distance from (ii)
| | Allow $k = \pm 4$ OR $\pm 4\sqrt{21}$ f.t. from (ii)
$\mathbf{b} = [2, -2, 3]$ | B1 | For stating correct direction
$\mathbf{a} = [-3, 13, 8]$ | A1 4 | For correct point seen in equation $\mathbf{r} = \mathbf{a} + l\mathbf{b}$
$\mathbf{r} = [-3, 13, 8] + l[2, -2, 3]$ | | AEF in this form

---