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(A) Given $\vec{a}+\vec{b}=m \vec{d}-\vec{c} \Rightarrow \vec{a}+\vec{b}+\vec{c}=m \vec{d} \quad (i)$ And $\vec{b}+\vec{c}=n \vec{a}-\vec{d} \Rightarr

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$\vec{a}-\vec{d}$

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(A) Given $\vec{a}+\vec{b}=m \vec{d}-\vec{c} \Rightarrow \vec{a}+\vec{b}+\vec{c}=m \vec{d} \quad (i)$ And $\vec{b}+\vec{c}=n \vec{a}-\vec{d} \Rightarrow \vec{d}=n \vec{a}-\vec{b}-\vec{c} \quad (ii)$ Substituting $(ii)$ into $(i)$,we get: $\vec{a}+\vec{b}+\vec{c}=m(n \vec{a}-\vec{b}-\vec{c})$ $\Rightarrow \vec{a}+\vec{b}+\vec{c}=mn \vec{a}-m \vec{b}-m \vec{c}$ $\Rightarrow (1-mn) \vec{a}+(1+m) \vec{b}+(1+m) \vec{c}=\vec{0}$ Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,they are linearly independent. Therefore,the coefficients must be zero: $1-mn=0 \Rightarrow mn=1$ and $1+m=0 \Rightarrow m=-1$. Substituting $m=-1$ into $mn=1$,we get $n=-1$. From $(i)$,$\vec{a}+\vec{b}+\vec{c}=(-1) \vec{d} \Rightarrow \vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$. Now,$3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d} = (\vec{a}+\vec{b}+\vec{c}+\vec{d}) + 2(\vec{a}+\vec{b}+\vec{c}) = \vec{0} + 2(-\vec{d}) + 2\vec{a} = 2\vec{a}-2\vec{d}$. Wait,re-evaluating the expression: $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d} = \vec{a} + 2(\vec{a}+\vec{b}+\vec{c}) + \vec{d} = \vec{a} + 2(-\vec{d}) + \vec{d} = \vec{a}-\vec{d}$.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$

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