Let vec{a}=2hat{i}-5hat{j}+5hat{k} and vec{b} | vedclass

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(C) Given $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$.This implies $(2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$, which means

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$218$

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(C) Given $2(\vec{a} 	imes \vec{c}) + 3(\vec{b} 	imes \vec{c}) = \vec{0}$.This implies $(2\vec{a} + 3\vec{b}) 	imes \vec{c} = \vec{0}$, which means $\vec{c}$ is parallel to $(2\vec{a} + 3\vec{b})$.Let $\vec{c} = \lambda(2\vec{a} + 3\vec{b})$.Calculating $2\vec{a} + 3\vec{b} = 2(2\hat{i} - 5\hat{j} + 5\hat{k}) + 3(\hat{i} - \hat{j} + 3\hat{k}) = (4+3)\hat{i} + (-10-3)\hat{j} + (10+9)\hat{k} = 7\hat{i} - 13\hat{j} + 19\hat{k}$.So, $\vec{c} = \lambda(7\hat{i} - 13\hat{j} + 19\hat{k})$.Given $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$.We have $\vec{a} - \vec{b} = (2-1)\hat{i} + (-5 - (-1))\hat{j} + (5-3)\hat{k} = \hat{i} - 4\hat{j} + 2\hat{k}$.Substituting into the dot product: $(\hat{i} - 4\hat{j} + 2\hat{k}) \cdot \lambda(7\hat{i} - 13\hat{j} + 19\hat{k}) = -97$.$\lambda(7 + 52 + 38) = -97 \Rightarrow 97\lambda = -97 \Rightarrow \lambda = -1$.Thus, $\vec{c} = -7\hat{i} + 13\hat{j} - 19\hat{k}$.Now, $\vec{c} 	imes \hat{k} = (-7\hat{i} + 13\hat{j} - 19\hat{k}) 	imes \hat{k} = -7(\hat{i} 	imes \hat{k}) + 13(\hat{j} 	imes \hat{k}) - 19(\hat{k} 	imes \hat{k}) = -7(-\hat{j}) + 13(\hat{i}) - 0 = 13\hat{i} + 7\hat{j}$.Finally, $|\vec{c} 	imes \hat{k}|^2 = 13^2 + 7^2 = 169 + 49 = 218$.

Let $\vec{a}=2\hat{i}-5\hat{j}+5\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3\hat{k}$. If $\vec{c}$ is a vector such that $2(\vec{a}\times\vec{c})+3(\vec{b}\times\vec{c})=\vec{0}$ and $(\vec{a}-\vec{b})\cdot\vec{c}=-97$, then $|\vec{c}\times \hat{k}|^{2}$ is equal to

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