Step 1: Understanding the Concept:
The notation $[\bar{a} \quad \bar{b} \quad \bar{c}]$ represents the scalar triple product of three vectors. This product is linear with respect to each of its vector arguments, allowing us to expand terms like $[x\bar{a} + y\bar{b} \quad \bar{c} \quad \bar{d}]$. It also has the property that swapping any two adjacent vectors changes the sign of the product.
Step 2: Key Formula or Approach:
1. Linearity: $[a\bar{u} + b\bar{v} \quad \bar{w} \quad \bar{x}] = a[\bar{u} \quad \bar{w} \quad \bar{x}] + b[\bar{v} \quad \bar{w} \quad \bar{x}]$.
2. Skew-symmetry: $[\bar{u} \quad \bar{v} \quad \bar{w}] = -[\bar{v} \quad \bar{u} \quad \bar{w}]$.
Expand the left-hand side (LHS) and rearrange it to match the terms on the right-hand side (RHS).
Step 3: Detailed Explanation:
Let's evaluate the LHS expression by expanding it using the linearity property:
LHS $= [2\bar{p} - 3\bar{r} \quad \bar{q} \quad \bar{s}] + [3\bar{p} + 2\bar{q} \quad \bar{r} \quad \bar{s}]$
Expand the first term:
$[2\bar{p} - 3\bar{r} \quad \bar{q} \quad \bar{s}] = [2\bar{p} \quad \bar{q} \quad \bar{s}] + [-3\bar{r} \quad \bar{q} \quad \bar{s}] = 2[\bar{p} \quad \bar{q} \quad \bar{s}] - 3[\bar{r} \quad \bar{q} \quad \bar{s}]$
Expand the second term:
$[3\bar{p} + 2\bar{q} \quad \bar{r} \quad \bar{s}] = [3\bar{p} \quad \bar{r} \quad \bar{s}] + [2\bar{q} \quad \bar{r} \quad \bar{s}] = 3[\bar{p} \quad \bar{r} \quad \bar{s}] + 2[\bar{q} \quad \bar{r} \quad \bar{s}]$
Now, add them together:
LHS $= 2[\bar{p} \quad \bar{q} \quad \bar{s}] - 3[\bar{r} \quad \bar{q} \quad \bar{s}] + 3[\bar{p} \quad \bar{r} \quad \bar{s}] + 2[\bar{q} \quad \bar{r} \quad \bar{s}]$
We need to format this to match the given RHS, which contains $[\bar{p} \quad \bar{r} \quad \bar{s}]$, $[\bar{q} \quad \bar{r} \quad \bar{s}]$, and $[\bar{p} \quad \bar{q} \quad \bar{s}]$.
Notice the term $-3[\bar{r} \quad \bar{q} \quad \bar{s}]$. We can swap $\bar{r}$ and $\bar{q}$ to make it match the required form, remembering that a swap introduces a negative sign:
$-3[\bar{r} \quad \bar{q} \quad \bar{s}] = -3(-[\bar{q} \quad \bar{r} \quad \bar{s}]) = +3[\bar{q} \quad \bar{r} \quad \bar{s}]$
Substitute this back into the LHS equation:
LHS $= 2[\bar{p} \quad \bar{q} \quad \bar{s}] + 3[\bar{q} \quad \bar{r} \quad \bar{s}] + 3[\bar{p} \quad \bar{r} \quad \bar{s}] + 2[\bar{q} \quad \bar{r} \quad \bar{s}]$
Group the like terms (the ones with $[\bar{q} \quad \bar{r} \quad \bar{s}]$):
LHS $= 3[\bar{p} \quad \bar{r} \quad \bar{s}] + (3+2)[\bar{q} \quad \bar{r} \quad \bar{s}] + 2[\bar{p} \quad \bar{q} \quad \bar{s}]$
LHS $= 3[\bar{p} \quad \bar{r} \quad \bar{s}] + 5[\bar{q} \quad \bar{r} \quad \bar{s}] + 2[\bar{p} \quad \bar{q} \quad \bar{s}]$
Now, compare this with the given RHS expression:
RHS $= m[\bar{p} \quad \bar{r} \quad \bar{s}] + n[\bar{q} \quad \bar{r} \quad \bar{s}] + t[\bar{p} \quad \bar{q} \quad \bar{s}]$
By comparing coefficients of corresponding scalar triple products, we get:
$m = 3$
$n = 5$
$t = 2$
Thus, the values are respectively 3, 5, 2.
Step 4: Final Answer:
The values of m, n, t are 3, 5, 2 respectively.