Let vec{a}=3 hat{i}+hat{j}-2 hat{k},vec{b}=4 | vedclass

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(D) Given $\vec{p} 	imes \vec{b} = \vec{c} 	imes \vec{b}$,we can write $\vec{p} 	imes \vec{b} - \vec{c} 	imes \vec{b} = \vec{0}$.This implies $(\v

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$32$

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(D) Given $\vec{p} 	imes \vec{b} = \vec{c} 	imes \vec{b}$,we can write $\vec{p} 	imes \vec{b} - \vec{c} 	imes \vec{b} = \vec{0}$.This implies $(\vec{p} - \vec{c}) 	imes \vec{b} = \vec{0}$.Therefore,$\vec{p} - \vec{c} = \lambda \vec{b}$,which means $\vec{p} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$.Given $\vec{p} \cdot \vec{a} = 0$,we substitute $\vec{p}$:$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0 \Rightarrow \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.Calculating the dot products:$\vec{c} \cdot \vec{a} = (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$.$\vec{b} \cdot \vec{a} = (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$.Substituting these values: $-8 + \lambda(-1) = 0 \Rightarrow \lambda = -8$.Thus,$\vec{p} = \vec{c} - 8 \vec{b} = (\hat{i} - 3 \hat{j} + 4 \hat{k}) - 8(4 \hat{i} + \hat{j} + 7 \hat{k}) = (1-32) \hat{i} + (-3-8) \hat{j} + (4-56) \hat{k} = -31 \hat{i} - 11 \hat{j} - 52 \hat{k}$.Finally,$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) = (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32$.

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