1. For $P$

Given volume of parallelepiped formed by $\vec a,\vec b,\vec c$ is $|\vec a\cdot(\vec b\times \vec c)|=2.$

We need volume formed by $2(\vec a\times \vec b),\quad 3(\vec b\times \vec c),\quad (\vec c\times \vec a).$

So required volume is $\left|2(\vec a\times \vec b)\cdot \left(3(\vec b\times \vec c)\times (\vec c\times \vec a)\right)\right|.$ But the standard identity gives: $[(\vec a\times \vec b),(\vec b\times \vec c),(\vec c\times \vec a)] = [\vec a,\vec b,\vec c]^2,$ where $[\vec a,\vec b,\vec c]=\vec a\cdot(\vec b\times \vec c)$.

Including scalar multipliers $2,3,1$: $\text{Required volume}=2\cdot 3\cdot 1\cdot [\vec a,\vec b,\vec c]^2=6\cdot 2^2=24.$

Thus, $P=24$ which matches List II (3).

2. For $Q$

Given $|[\vec a,\vec b,\vec c]|=5.$

We need volume of parallelepiped formed by $3(\vec a+\vec b),\quad (\vec b+\vec c),\quad 2(\vec c+\vec a).$

Hence volume is $|3\cdot 1\cdot 2|\, |[(\vec a+\vec b), (\vec b+\vec c), (\vec c+\vec a)]|.$ So, $V=6\, |[(\vec a+\vec b), (\vec b+\vec c), (\vec c+\vec a)]|.$

Now expand the scalar triple product linearly: $[(\vec a+\vec b), (\vec b+\vec c), (\vec c+\vec a)]$ $=[\vec a,\vec b,\vec c]+[\vec a,\vec b,\vec a]+[\vec a,\vec c,\vec c]+[\vec a,\vec c,\vec a]$ $\quad +[\vec b,\vec b,\vec c]+[\vec b,\vec b,\vec a]+[\vec b,\vec c,\vec c]+[\vec b,\vec c,\vec a].$

All terms having two equal vectors are zero. So only $[\vec a,\vec b,\vec c]+[\vec b,\vec c,\vec a]$ remain. These are equal cyclic permutations, hence $[(\vec a+\vec b), (\vec b+\vec c), (\vec c+\vec a)] = 2[\vec a,\vec b,\vec c].$

Therefore, $V=6\cdot 2\cdot 5=60.$

Thus, $Q=60$ which matches List II (4).

3. For $R$

Area of triangle with adjacent sides $\vec a,\vec b$ is $20$.

Since triangle area is half the parallelogram area, $\frac12 |\vec a\times \vec b|=20 \implies |\vec a\times \vec b|=40.$

We need area of triangle with adjacent sides $2\vec a+3\vec b,\quad \vec a-\vec b.$

First compute cross product: $(2\vec a+3\vec b)\times(\vec a-\vec b)$ $=2\vec a\times \vec a -2\vec a\times \vec b +3\vec b\times \vec a -3\vec b\times \vec b.$ Since $\vec a\times \vec a=0,\quad \vec b\times \vec b=0,\quad \vec b\times \vec a=-\vec a\times \vec b,$ we get $(2\vec a+3\vec b)\times(\vec a-\vec b) = -2\vec a\times \vec b -3\vec a\times \vec b = -5\vec a\times \vec b.$ Hence $\left|(2\vec a+3\vec b)\times(\vec a-\vec b)\right|=5|\vec a\times \vec b|=5\cdot 40=200.$

Therefore required triangle area is $\frac12\cdot 200=100.$

Thus, $R=100$ which matches List II (1).

4. For $S$

Area of parallelogram with adjacent sides $\vec a,\vec b$ is $30$, so $|\vec a\times \vec b|=30.$

We need area of parallelogram with adjacent sides $(\vec a+\vec b),\quad \vec a.$

Its area is $|(\vec a+\vec b)\times \vec a|=|\vec a\times \vec a + \vec b\times \vec a|.$ Since $\vec a\times \vec a=0$ and $\vec b\times \vec a=-\vec a\times \vec b$, $|(\vec a+\vec b)\times \vec a|=|\vec b\times \vec a|=|\vec a\times \vec b|=30.$

Thus, $S=30$ which matches List II (2).

5. Final matching

We have:

• $P=24 \Rightarrow (3)$
• $Q=60 \Rightarrow (4)$
• $R=100 \Rightarrow (1)$
• $S=30 \Rightarrow (2)$

So the correct code is $P=3,\quad Q=4,\quad R=1,\quad S=2.$

This corresponds to Option C.