If \([2 \overline{\mathrm{p}}-3 \overline{\mathrm{r}} ~\overline{\mathrm{q}} ~\overline{\mathrm{s}}]+[3 \overline{\mathrm{p}}+2 \overline{\mathrm{q}} ~\overline{\mathrm{r}} ~\overline{\mathrm{s}}]=\mathrm{m}[\overline{\mathrm{p}} ~\overline{\mathrm{r}} ~\overline{\mathrm{s}}]+\mathrm{n}[\overline{\mathrm{q}} ~\overline{\mathrm{r}} ~\overline{\mathrm{s}}]+\mathrm{t}[\overline{\mathrm{p}} ~\overline{\mathrm{q}} ~\overline{\mathrm{s}}]\),, then the values of $m, n, t$ respectively are ....

Given: $[2\vec{p} - 3\vec{r} \ \vec{q} \ \vec{s}] + [3\vec{p} + 2\vec{q} \ \vec{r} \ \vec{s}] = m[\vec{p} \ \vec{r} \ \vec{s}] + n[\vec{q} \ \vec{r} \ \vec{s}] + t[\vec{p} \ \vec{q} \ \vec{s}]$

Using linearity of scalar triple products: $[\vec{a} + \vec{b} \ \vec{c} \ \vec{d}] = [\vec{a} \ \vec{c} \ \vec{d}] + [\vec{b} \ \vec{c} \ \vec{d}]$. Expand the first term:
$[2\vec{p} - 3\vec{r} \ \vec{q} \ \vec{s}] = 2[\vec{p} \ \vec{q} \ \vec{s}] - 3[\vec{r} \ \vec{q} \ \vec{s}]$
Since $[\vec{r} \ \vec{q} \ \vec{s}] = -[\vec{q} \ \vec{r} \ \vec{s}]$, this simplifies to $2[\vec{p} \ \vec{q} \ \vec{s}] + 3[\vec{q} \ \vec{r} \ \vec{s}]$.

Similarly, the second term expands to $3[\vec{p} \ \vec{r} \ \vec{s}] + 2[\vec{q} \ \vec{r} \ \vec{s}]$.

Combining both terms gives the left-hand side:
$\textbf{LHS} = 3[\vec{p} \ \vec{r} \ \vec{s}] + 5[\vec{q} \ \vec{r} \ \vec{s}] + 2[\vec{p} \ \vec{q} \ \vec{s}]$

Matching coefficients with the right-hand side yields $m = 3$, $n = 5$, $t = 2$.