If the vectors vec{b}, vec{c}, vec{d} are not | vedclass

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(A) Let $\vec{x} = (\vec{a} 	imes \vec{b}) 	imes(\vec{c} 	imes \vec{d}) + (\vec{a} 	imes \vec{c}) 	imes(\vec{d} 	imes \vec{b}) + (\vec{a} 	imes

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parallel to $\vec{a}$

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(A) Let $\vec{x} = (\vec{a} 	imes \vec{b}) 	imes(\vec{c} 	imes \vec{d}) + (\vec{a} 	imes \vec{c}) 	imes(\vec{d} 	imes \vec{b}) + (\vec{a} 	imes \vec{d}) 	imes(\vec{b} 	imes \vec{c})$.Using the vector identity $(\vec{p} 	imes \vec{q}) 	imes \vec{r} = (\vec{p} \cdot \vec{r})\vec{q} - (\vec{q} \cdot \vec{r})\vec{p}$, we expand each term:$1$. $(\vec{a} 	imes \vec{b}) 	imes(\vec{c} 	imes \vec{d}) = [\vec{a} \vec{c} \vec{d}]\vec{b} - [\vec{b} \vec{c} \vec{d}]\vec{a}$$2$. $(\vec{a} 	imes \vec{c}) 	imes(\vec{d} 	imes \vec{b}) = [\vec{a} \vec{d} \vec{b}]\vec{c} - [\vec{c} \vec{d} \vec{b}]\vec{a}$$3$. $(\vec{a} 	imes \vec{d}) 	imes(\vec{b} 	imes \vec{c}) = [\vec{a} \vec{b} \vec{c}]\vec{d} - [\vec{d} \vec{b} \vec{c}]\vec{a}$Summing these, we get:$\vec{x} = ([\vec{a} \vec{c} \vec{d}]\vec{b} + [\vec{a} \vec{d} \vec{b}]\vec{c} + [\vec{a} \vec{b} \vec{c}]\vec{d}) - ([\vec{b} \vec{c} \vec{d}] + [\vec{c} \vec{d} \vec{b}] + [\vec{d} \vec{b} \vec{c}])\vec{a}$Since the scalar triple product is cyclic, $[\vec{b} \vec{c} \vec{d}] = [\vec{c} \vec{d} \vec{b}] = [\vec{d} \vec{b} \vec{c}]$.Thus, the coefficient of $\vec{a}$ is $-3[\vec{b} \vec{c} \vec{d}]$.By the property of scalar triple products, $[\vec{a} \vec{c} \vec{d}]\vec{b} + [\vec{a} \vec{d} \vec{b}]\vec{c} + [\vec{a} \vec{b} \vec{c}]\vec{d} = [\vec{b} \vec{c} \vec{d}]\vec{a}$.Substituting this back, $\vec{x} = [\vec{b} \vec{c} \vec{d}]\vec{a} - 3[\vec{b} \vec{c} \vec{d}]\vec{a} = -2[\vec{b} \vec{c} \vec{d}]\vec{a}$.Since $\vec{x}$ is a scalar multiple of $\vec{a}$, it is parallel to $\vec{a}$.

If the vectors $\vec{b}, \vec{c}, \vec{d}$ are not coplanar, then the vector $(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})+(\vec{a} \times \vec{c}) \times(\vec{d} \times \vec{b})+(\vec{a} \times \vec{d}) \times(\vec{b} \times \vec{c})$ is

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