If [vec{p}-vec{r}, vec{q}, vec{s}] + [vec{p}+ | vedclass

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(D) The scalar triple product is defined as $[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.Using the linearity property of the

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$(1, 1, 1)$

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(D) The scalar triple product is defined as $[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.Using the linearity property of the scalar triple product:$[\vec{p}-\vec{r}, \vec{q}, \vec{s}] = [\vec{p}, \vec{q}, \vec{s}] - [\vec{r}, \vec{q}, \vec{s}] = [\vec{p}, \vec{q}, \vec{s}] + [\vec{q}, \vec{r}, \vec{s}]$.$[\vec{p}+\vec{q}, \vec{r}, \vec{s}] = [\vec{p}, \vec{r}, \vec{s}] + [\vec{q}, \vec{r}, \vec{s}]$.Adding these two expressions:$([\vec{p}, \vec{q}, \vec{s}] + [\vec{q}, \vec{r}, \vec{s}]) + ([\vec{p}, \vec{r}, \vec{s}] + [\vec{q}, \vec{r}, \vec{s}]) = [\vec{p}, \vec{r}, \vec{s}] + 2[\vec{q}, \vec{r}, \vec{s}] + [\vec{p}, \vec{q}, \vec{s}]$.Comparing this with $m[\vec{p}, \vec{r}, \vec{s}] + n[\vec{q}, \vec{r}, \vec{s}] + t[\vec{p}, \vec{q}, \vec{s}]$,we get $m=1$,$n=2$,$t=1$.Since the provided options do not contain $(1, 2, 1)$,we correct the option $D$ to $(1, 2, 1)$.

If $[\vec{p}-\vec{r}, \vec{q}, \vec{s}] + [\vec{p}+\vec{q}, \vec{r}, \vec{s}] = m[\vec{p}, \vec{r}, \vec{s}] + n[\vec{q}, \vec{r}, \vec{s}] + t[\vec{p}, \vec{q}, \vec{s}]$,then the values of $m$,$n$,$t$ respectively are . . . . . .

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