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(A) Let $\vec{N_1}$ and $\vec{N_2}$ be the normal vectors to the planes $P_1$ and $P_2$ respectively. Since $P_1$ is determined by $\vec{a}$ and $\vec

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(A) Let $\vec{N_1}$ and $\vec{N_2}$ be the normal vectors to the planes $P_1$ and $P_2$ respectively. Since $P_1$ is determined by $\vec{a}$ and $\vec{b}$, its normal is $\vec{N_1} = \vec{a} 	imes \vec{b}$. Since $P_2$ is determined by $\vec{c}$ and $\vec{d}$, its normal is $\vec{N_2} = \vec{c} 	imes \vec{d}$. Given that $(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) = \vec{0}$, we have $\vec{N_1} 	imes \vec{N_2} = \vec{0}$. This implies that the normal vectors $\vec{N_1}$ and $\vec{N_2}$ are parallel to each other. Since the normal vectors are parallel, the planes $P_1$ and $P_2$ are parallel. Therefore, the angle between the planes $P_1$ and $P_2$ is $0$.

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

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