Matching Vectors to Volumes and Areas of Geometric Figures

To find the correct matching between List - I and List - II, let us evaluate each item step-by-step:

Item P:

Let the volume of the parallelepiped determined by vectors $\vec{a}, \vec{b}, \vec{c}$ be: $V = |[\vec{a}\ \vec{b}\ \vec{c}]| = 2$

The volume of the parallelepiped formed by the vectors $2(\vec{a} \times \vec{b})$, $3(\vec{b} \times \vec{c})$, and $(\vec{c} \times \vec{a})$ is given by the scalar triple product: $V_P = |[2(\vec{a} \times \vec{b})\quad 3(\vec{b} \times \vec{c})\quad (\vec{c} \times \vec{a})]|$

Using the properties of the scalar triple product: $V_P = (2 \times 3 \times 1) \cdot |[\vec{a} \times \vec{b}\quad \vec{b} \times \vec{c}\quad \vec{c} \times \vec{a}]|$

We know the identity: $[\vec{a} \times \vec{b}\quad \vec{b} \times \vec{c}\quad \vec{c} \times \vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2$

Substituting the values: $V_P = 6 \cdot |[\vec{a}\ \vec{b}\ \vec{c}]|^2 = 6 \cdot (2)^2 = 6 \cdot 4 = 24$

Thus, $\text{P} \rightarrow 3$.

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Item Q:

Let the volume of the parallelepiped determined by vectors $\vec{a}, \vec{b}, \vec{c}$ be: $V = |[\vec{a}\ \vec{b}\ \vec{c}]| = 5$

The volume of the parallelepiped formed by $3(\vec{a} + \vec{b})$, $(\vec{b} + \vec{c})$, and $2(\vec{c} + \vec{a})$ is: $V_Q = |[3(\vec{a} + \vec{b})\quad (\vec{b} + \vec{c})\quad 2(\vec{c} + \vec{a})]|$ $V_Q = (3 \times 1 \times 2) \cdot |[\vec{a} + \vec{b}\quad \vec{b} + \vec{c}\quad \vec{c} + \vec{a}]|$

Using the identity: $[\vec{a} + \vec{b}\quad \vec{b} + \vec{c}\quad \vec{c} + \vec{a}] = 2[\vec{a}\ \vec{b}\ \vec{c}]$

We get: $V_Q = 6 \cdot 2 \cdot |[\vec{a}\ \vec{b}\ \vec{c}]| = 12 \cdot 5 = 60$

Thus, $\text{Q} \rightarrow 4$.

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Item R:

The area of a triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is: $\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 20 \implies |\vec{a} \times \vec{b}| = 40$

The area of the triangle with adjacent sides $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is: $A_R = \frac{1}{2} |(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b})|$

Evaluating the 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 $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$: $(2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b}) = -2(\vec{a} \times \vec{b}) - 3(\vec{a} \times \vec{b}) = -5(\vec{a} \times \vec{b})$

Therefore: $A_R = \frac{1}{2} |-5(\vec{a} \times \vec{b})| = \frac{5}{2} |\vec{a} \times \vec{b}| = \frac{5}{2} \times 40 = 100$

Thus, $\text{R} \rightarrow 1$.

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Item S:

The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is: $\text{Area} = |\vec{a} \times \vec{b}| = 30$

The area of the parallelogram with adjacent sides $(\vec{a} + \vec{b})$ and $\vec{a}$ is: $A_S = |(\vec{a} + \vec{b}) \times \vec{a}| = |\vec{a} \times \vec{a} + \vec{b} \times \vec{a}| = |-\vec{a} \times \vec{b}| = |\vec{a} \times \vec{b}| = 30$

Thus, $\text{S} \rightarrow 2$.

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Conclusion:

Matching List - I with List - II: $\text{P} \rightarrow 3,\ \text{Q} \rightarrow 4,\ \text{R} \rightarrow 1,\ \text{S} \rightarrow 2$

Therefore, the correct option is C.