In triangle $ABC$ shown below, sides $AB$ and $AC$ are $(p - 6\sqrt{2})\text{ cm}$ and $(4 - \sqrt{2})\text{ cm}$ respectively. Angle $BAC = 45^\circ$.

1. Set up the equation for the area of the triangle using the formula $\frac{1}{2}ab\sin C$.

$\text{Area} = \frac{1}{2}(AB)(AC)\sin(\angle BAC)$
$\frac{1}{2}(p - 6\sqrt{2})(4 - \sqrt{2})\sin 45^\circ = 17\sqrt{2} - 19$

2. Substitute the exact value of $\sin 45^\circ$ and expand the brackets.

$\frac{1}{2}(p - 6\sqrt{2})(4 - \sqrt{2})\left(\frac{\sqrt{2}}{2}\right) = 17\sqrt{2} - 19$
$\frac{\sqrt{2}}{4}(4p - p\sqrt{2} - 24\sqrt{2} + 12) = 17\sqrt{2} - 19$

3. Distribute $\frac{\sqrt{2}}{4}$ into the expanded expression.

$\frac{4p\sqrt{2} - 2p - 48 + 12\sqrt{2}}{4} = 17\sqrt{2} - 19$
$p\sqrt{2} - \frac{1}{2}p - 12 + 3\sqrt{2} = 17\sqrt{2} - 19$
$(p + 3)\sqrt{2} - \left(\frac{1}{2}p + 12\right) = 17\sqrt{2} - 19$

4. Equate the rational or irrational parts to solve for $p$.

Comparing the coefficients of $\sqrt{2}$:
$p + 3 = 17$
$p = 14$

(Alternatively, comparing the rational parts:
$-\frac{1}{2}p - 12 = -19 \implies -\frac{1}{2}p = -7 \implies p = 14$)

1. Express the perpendicular distance in terms of $AB$ and $\sin 45^\circ$

Let the perpendicular distance from $B$ to $AC$ be $h$.
$h = AB \sin 45^\circ$

2. Substitute the known values into the expression

From part (i), $p = 14$, so $AB = 14 - 6\sqrt{2}$.
$\sin 45^\circ = \frac{\sqrt{2}}{2}$
$h = (14 - 6\sqrt{2})\left(\frac{\sqrt{2}}{2}\right)$

3. Expand and simplify the surd expression

$h = \frac{14\sqrt{2}}{2} - \frac{6(\sqrt{2})(\sqrt{2})}{2}$
$h = 7\sqrt{2} - \frac{6(2)}{2}$
$h = 7\sqrt{2} - 6$