Simplification of an expression containing $\operatorname{Li}_3(x)$ terms

Surprisingly, $\mathcal K$ can be expressed in elementary terms. Let, $$a = \ln 2\\ b=\ln 3\\ c=\ln 5$$ Then, $$\mathcal{K}=\frac23(878 a^3 - 37 b^3 - 7 c^3) - 2 a^2 (202 b + 133 c) + 4 b^2 (-32 a + 19 c) + 3 c^2(13 a - 21 b) + 278 a b c - \frac23 \pi^2 (22 a - 50 b + 25 c) \approx -7.809651$$

Too long for a comment, but a related thing, really far from the solution.

$$\operatorname{Li}_3\left(\frac{3}{8}\right)+\operatorname{Li}_3\left(\frac{5}{8}\right)-\operatorname{Li}_3\left(\frac{3}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{8}\right)-\frac{1}{2}\ln\left(\frac{3}{8}\right)\ln^2\left(\frac{5}{8}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{8}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{9}{25}\right).$$

I get it by using this identitiy for $z:=3/8$.

An other one for $z:=1/6$ with the same identity.

$$\operatorname{Li}_3\left(\frac{1}{6}\right)+\operatorname{Li}_3\left(\frac{5}{6}\right)-\operatorname{Li}_3\left(\frac{1}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{6}\right)-\frac{1}{2}\ln\left(\frac{1}{6}\right)\ln^2\left(\frac{5}{6}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{6}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{1}{25}\right).$$

$\def\tfrac#1#2{\textstyle\frac{#1}{#2}}$I found (numerically, using the PSLQ algorithm) only five vanishing rational relations between the fifteen trilogarithms $$ L(\tfrac uv) = \mathrm{Li}_3(\tfrac uv), \qquad 0<u<v,\quad v\in\{2,3,4,5,6,8\}.$$ The first three are well-known: $$ \begin{eqnarray} 0 &=& - L(\tfrac12) + \tfrac78\zeta (3)-\tfrac{1}{2} \zeta (2) \log2+\tfrac16\log^32\\ 0 &=& 6 L(\tfrac{3}{4})+12 L(\tfrac{1}{3})+6 L(\tfrac{1}{4}) \\&&-19 \zeta (3)+12 \zeta (2) \log2-2 \log^33-16 \log^32+12 \log^22 \log3\\ 0 &=&-24 L(\tfrac{2}{3})-24 L(\tfrac{1}{3})-6 L(\tfrac{1}{4})\\&&+45 \zeta (3)-24 \zeta (2) \log3+12 \zeta (2) \log2+8 \log^33+8 \log^32-12 \log2 \log^23 \end{eqnarray} $$ These two, I think, might not be: $$\begin{eqnarray} 0&=& 12 L(\tfrac{4}{5})-42 L(\tfrac{2}{3})+12 L(\tfrac{3}{5})+12 L(\tfrac{2}{5})+6 L(\tfrac{3}{8})-36 L(\tfrac{1}{3})+12 L(\tfrac{1}{5})+6 L(\tfrac{1}{6}) \\&&+19 \zeta (3)+24 \zeta (2) \log^25-48 \zeta (2) \log3+12 \zeta (2) \log2-8 \log^35+12 \log^33-28 \log^32+6 \log3 \log^25+18 \log2 \log^25-24 \log2 \log^23+24 \log^22 \log3-12 \log2 \log3 \log^25 \\0&=& 48 L(\tfrac{5}{6})+36 L(\tfrac{4}{5})-144 L(\tfrac{2}{3})-24 L(\tfrac{5}{8})+72 L(\tfrac{2}{5})-144 L(\tfrac{1}{3})+60 L(\tfrac{1}{5})+48 L(\tfrac{1}{6}) \\&&+63 \zeta (3)+48 \zeta (2) \log^25-96 \zeta (2) \log3+12 \zeta (2) \log2-28 \log^35+32 \log^33+92 \log^32+72 \log2 \log^25+24 \log^23 \log^25-84 \log^22 \log^25-120 \log2 \log^23-48 \log^22 \log3+48 \log2 \log3 \log^25 \end{eqnarray}$$

A linear combination $$0=-\tfrac{21}{2}\mathrm{I}+\tfrac{19}{6}\mathrm{II}$$ of the last two relations just happens to eliminate all the trilogarithm terms from your expression, giving the following expression equal to yours: $$ -100 \zeta (2) \log^25+200 \zeta (2) \log3-88 \zeta (2) \log2-\tfrac{14}{3} \log^35-\tfrac{74}{3} \log^33\\+\tfrac{1756}{3} \log^32-63 \log3 \log^25 +39 \log2 \log^25+76 \log^23 \log^25\\-266 \log^22 \log^25-128 \log2 \log^23-404 \log^22 \log3+278 \log2 \log3 \log^25 $$

Here are the identities in a more computer-readable form:

{{-6, 6, -1, 1, -12, 6, 3, 3}.{Log[2]*Log[2]*Log[2], Log[2]*Log[2]*Log[3], Log[3]*Log[3]*Log[3], Zeta[3], PolyLog[3, 1/2], PolyLog[3, 1/3], PolyLog[3, 1/4], PolyLog[3, 3/4]}, {2, -6, 4, 12, -12, 12, 12, -12, -12, -3}.{Log[2]*Log[2]*Log[2], Log[2]*Log[3]*Log[3], Log[3]*Log[3]*Log[3], Zeta[2]*Log[2], Zeta[2]*Log[3], Zeta[3], PolyLog[3, 1/2], PolyLog[3, 1/3], PolyLog[3, 2/3], PolyLog[3, 1/4]}, {-28, 24, -24, -12, 18, 12, 6, -8, 12, -48, 24, 19, -36, -42, 12, 12, 12, 12, 6, 6}.{Log[2]*Log[2]*Log[2], Log[2]*Log[2]*Log[3], Log[2]*Log[3]*Log[3], Log[2]*Log[3]*Log[5], Log[2]*Log[5]*Log[5], Log[3]*Log[3]*Log[3], Log[3]*Log[5]*Log[5], Log[5]*Log[5]*Log[5], Zeta[2]*Log[2], Zeta[2]*Log[3], Zeta[2]*Log[5], Zeta[3], PolyLog[3, 1/3], PolyLog[3, 2/3], PolyLog[3, 1/5], PolyLog[3, 2/5], PolyLog[3, 3/5], PolyLog[3, 4/5], PolyLog[3, 1/6], PolyLog[3, 3/8]}, {92, -48, -84, -120, 48, 72, 32, 24, -28, 12, -96, 48, 63, -144, -144, 60, 72, 36, 48, 48, -24}.{Log[2]*Log[2]*Log[2], Log[2]*Log[2]*Log[3], Log[2]*Log[2]*Log[5], Log[2]*Log[3]*Log[3], Log[2]*Log[3]*Log[5], Log[2]*Log[5]*Log[5], Log[3]*Log[3]*Log[3], Log[3]*Log[3]*Log[5], Log[5]*Log[5]*Log[5], Zeta[2]*Log[2], Zeta[2]*Log[3], Zeta[2]*Log[5], Zeta[3], PolyLog[3, 1/3], PolyLog[3, 2/3], PolyLog[3, 1/5], PolyLog[3, 2/5], PolyLog[3, 4/5], PolyLog[3, 1/6], PolyLog[3, 5/6], PolyLog[3, 5/8]}}

and here is the final expression:

{1756, -1212, -798, -384, 834, 117, -74, 228, -189, -14, -264, 600, -300}.{Log[2]*Log[2]*Log[2], Log[2]*Log[2]*Log[3],Log[2]*Log[2]*Log[5], Log[2]*Log[3]*Log[3], Log[2]*Log[3]*Log[5], Log[2]*Log[5]*Log[5], Log[3]*Log[3]*Log[3], Log[3]*Log[3]*Log[5], Log[3]*Log[5]*Log[5], Log[5]*Log[5]*Log[5], Log[2]*Zeta[2], Log[3]*Zeta[2], Log[5]*Zeta[2]}/(3)