Let the plane P: vec{r} cdot vec{a} = d conta | vedclass

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(B) The equation of a plane passing through the line of intersection of two planes $P_1: \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 6 = 0$ and $P_

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$93$

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(B) The equation of a plane passing through the line of intersection of two planes $P_1: \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 6 = 0$ and $P_2: \vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) - 7 = 0$ is given by $P_1 + \lambda P_2 = 0$.Substituting the given planes,we get:$(\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) - 6) + \lambda(\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) - 7) = 0$Since the plane passes through the point $(2, 3, 1/2)$,we substitute $\vec{r} = 2\hat{i} + 3\hat{j} + \frac{1}{2}\hat{k}$:$(2 + 9 - 1/2 - 6) + \lambda(-12 + 15 - 1/2 - 7) = 0$$(11 - 6.5) + \lambda(3 - 7.5) = 0$$4.5 - 4.5\lambda = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ back into the equation:$\vec{r} \cdot ((\hat{i} - 6\hat{i}) + (3\hat{j} + 5\hat{j}) + (-\hat{k} - \hat{k})) = 6 + 7$$\vec{r} \cdot (-5\hat{i} + 8\hat{j} - 2\hat{k}) = 13$.Comparing this with $\vec{r} \cdot \vec{a} = d$,we have $\vec{a} = -5\hat{i} + 8\hat{j} - 2\hat{k}$ and $d = 13$.Then $|\vec{a}|^2 = (-5)^2 + 8^2 + (-2)^2 = 25 + 64 + 4 = 93$.We need to find $\frac{|13\vec{a}|^2}{d^2} = \frac{13^2 |\vec{a}|^2}{d^2} = \frac{169 	imes 93}{169} = 93$.

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