Let vec{a} = hat{i} + hat{j} + hat{k}, vec{b} | vedclass

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(C) Given $\vec{p} 	imes \vec{a} = \vec{b} 	imes \vec{a}$, we can write $\vec{p} 	imes \vec{a} - \vec{b} 	imes \vec{a} = 0$, which implies $(\vec{

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$\sqrt{17}$

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(C) Given $\vec{p} 	imes \vec{a} = \vec{b} 	imes \vec{a}$, we can write $\vec{p} 	imes \vec{a} - \vec{b} 	imes \vec{a} = 0$, which implies $(\vec{p} - \vec{b}) 	imes \vec{a} = 0$.This means $(\vec{p} - \vec{b})$ is parallel to $\vec{a}$, so $\vec{p} - \vec{b} = t\vec{a}$ for some scalar $t$.Thus, $\vec{p} = \vec{b} + t\vec{a} = (\hat{i} - 3\hat{j} + 2\hat{k}) + t(\hat{i} + \hat{j} + \hat{k}) = (1+t)\hat{i} + (t-3)\hat{j} + (t+2)\hat{k}$.Given $\vec{p} \cdot \vec{c} = 0$, where $\vec{c} = 3\hat{i} - 2\hat{k}$, we have $((1+t)\hat{i} + (t-3)\hat{j} + (t+2)\hat{k}) \cdot (3\hat{i} - 2\hat{k}) = 0$.$3(1+t) - 2(t+2) = 0 \implies 3 + 3t - 2t - 4 = 0 \implies t - 1 = 0 \implies t = 1$.Substituting $t=1$ into $\vec{p}$, we get $\vec{p} = (1+1)\hat{i} + (1-3)\hat{j} + (1+2)\hat{k} = 2\hat{i} - 2\hat{j} + 3\hat{k}$.Finally, $|\vec{p}| = \sqrt{2^2 + (-2)^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} - 2\hat{k}$. If a vector $\vec{p}$ satisfies the conditions $\vec{p} \cdot \vec{c} = 0$ and $\vec{p} \times \vec{a} = \vec{b} \times \vec{a}$, then the value of $|\vec{p}|$ is:

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