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(D) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} 	imes \vec{b}| = 15$. Now,consider the parallelogra

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$120$

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(D) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} 	imes \vec{b}| = 15$. Now,consider the parallelogram with adjacent sides $(3 \vec{a} + \vec{b})$ and $(\vec{a} + 3 \vec{b})$. The area of this parallelogram is $|(3 \vec{a} + \vec{b}) 	imes (\vec{a} + 3 \vec{b})|$. Expanding the cross product: $|(3 \vec{a} + \vec{b}) 	imes (\vec{a} + 3 \vec{b})| = |3 \vec{a} 	imes \vec{a} + 9 \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{a} + 3 \vec{b} 	imes \vec{b}|$. Since $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{b} = 0$,and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$,we have: $|0 + 9(\vec{a} 	imes \vec{b}) - (\vec{a} 	imes \vec{b}) + 0| = |8(\vec{a} 	imes \vec{b})| = 8 |\vec{a} 	imes \vec{b}|$. Substituting the given area $|\vec{a} 	imes \vec{b}| = 15$: Area $= 8 	imes 15 = 120$ sq. units.

If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ sq. units,then the area of the parallelogram having $3 \vec{a}+\vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides,in square units,is

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