Maybe that's too much work

The formula for the radius $r$ of the incenter is $r = \dfrac{A}{s}$ , where $A$ is the area of the triangle and $s$ is the semi-perimeter.

Using Heron's formula $A = \sqrt{s(s - a)(s - b)(s - c)}$ where $a,b,c$ are the side lengths, we find that

$A = \sqrt{(\frac{21}{2})(\frac{9}{2})(\frac{7}{2})(\frac{5}{2})} = \dfrac{21\sqrt{15}}{4}.$

Thus $r = \dfrac{\frac{21\sqrt{15}}{4}}{\frac{21}{2}} = \dfrac{\sqrt{15}}{2}.$

Now draw an altitude from $A$ to $BC$ . Let this altitude intersect $PQ$ at $M$ and $BC$ at $N$ . Then

$\dfrac{PQ}{BC} = \dfrac{AM}{AN} \Longrightarrow PQ = 6\dfrac{(AN - r)}{AN}.$

So now we need to find $AN$ . The Cosine rule gives us that

$7^{2} = 6^{2} + 8^{2} - 2(6)(8)\cos(\angle B) \Longrightarrow \cos(\angle B) = \dfrac{51}{96} = \dfrac{17}{32}$

$\sin(\angle B) = \sqrt{1 - (\frac{17}{32})^{2}} = \dfrac{7\sqrt{15}}{32}.$

Thus $AN = 8\sin(\angle B) = \dfrac{7\sqrt{15}}{4}$ , and so

$PQ = 6\dfrac{\frac{7\sqrt{15}}{4} - \frac{\sqrt{15}}{2}}{\frac{7\sqrt{15}}{4}} = \dfrac{30}{7}.$

So now we need to find the $30$ th root of $30^{7}$ as expressed in base $7$ . Using WolframAlpha we have that $30^{7}$ in base $7$ is $1402646634642$ , the $30$ th root of which is $2.5403776....$ , the least integer greater than this being $\boxed{3}$ .

(I assumed that the $30$ th root was to be taken as if $30^{7}$ in base $7$ was a base $10$ number. I found the wording a bit confusing. @Azhaghu Roopesh M Is this what you had in mind once the value $PQ = \frac{30}{7}$ was found?)

My solution was similar to Brian's up to the point of the triangle similarity. I worked it like this:

Firstly, I will find the radius of the incircle. I will do this by finding the area $S$ of the triangle:

Let $s$ be the semiperimeter of the triangle. Then, $s = \frac{21}{2}$ . Thus:

$S = \sqrt{\frac{21}{2}\frac{5}{2}\frac{7}{2}\frac{9}{2}} = \frac{21\sqrt{15}}{4}$

Now, the radius $r$ can be found using the formula: $S = s*r$

Therefore, $r*\frac{21}{2} = \frac{21\sqrt{15}}{4} \rightarrow r = \frac{\sqrt{15}}{2}$ .

Now, we find the distance from the opposite vertex to the smallest side. This can be done by using areas once more: $\frac{6*d}{2} = \frac{21\sqrt{15}}{4}$

Thus, $d = \frac{7\sqrt{15}}{4}$

Now, using triangle similarity, let $\overline{PQ} = x$ . Then: $\frac{d - r}{x} = \frac{d}{6}$ .

$\frac{\frac{5\sqrt{15}}{4}}{x} = \frac{\frac{7\sqrt{15}}{4}}{6}$

$\frac{5}{x} = \frac{7}{6}$

$x = \frac{30}{7}$

Now, $a^{z} = 30^{7}$ in base $10$ ; $30_{10} = 42_{7}$ , and then in base $7$ we have $a^{z} = 1402646634642_{7}$

The $30$ -th root of the number $1402646634642_{10}$ is approximately $2.54$ , and thus the number sought is 3.