Let a plane P pass through the point A(3, 7, | vedclass

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(B) The line $L$ passes through point $B(2, 3, -2)$ and has direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$.The vector $\vec{AB} = (2-3)\h

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$3$

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(B) The line $L$ passes through point $B(2, 3, -2)$ and has direction vector $\vec{v} = -3\hat{i} + 2\hat{j} + \hat{k}$.The vector $\vec{AB} = (2-3)\hat{i} + (3-7)\hat{j} + (-2 - (-7))\hat{k} = -\hat{i} - 4\hat{j} + 5\hat{k}$.The normal vector $\vec{n}$ to the plane is $\vec{n} = \vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -4 & 5 \ -3 & 2 & 1 \end{vmatrix} = \hat{i}(-4-10) - \hat{j}(-1+15) + \hat{k}(-2-12) = -14\hat{i} - 14\hat{j} - 14\hat{k}$.We can take $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The equation of the plane is $1(x-3) + 1(y-7) + 1(z+7) = 0$, which simplifies to $x + y + z - 3 = 0$.The distance $d$ from the origin $(0, 0, 0)$ is $d = \frac{|0+0+0-3|}{\sqrt{1^2+1^2+1^2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.Therefore, $d^2 = 3$.

Let a plane $P$ pass through the point $A(3, 7, -7)$ and contain the line $L: \frac{x - 2}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$. If the distance of the plane $P$ from the origin is $d$, then $d^2$ is

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