Let A=(-3,-2,7) and B=(3,1,-2). Let a plane p | vedclass

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(D) Let the plane $P$ divide the line segment $AB$ at point $Q$ in the ratio $2:1$. Using the section formula,the coordinates of $Q$ are: $Q = \left(

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$-1$

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(D) Let the plane $P$ divide the line segment $AB$ at point $Q$ in the ratio $2:1$. Using the section formula,the coordinates of $Q$ are: $Q = \left( \frac{2(3) + 1(-3)}{2+1}, \frac{2(1) + 1(-2)}{2+1}, \frac{2(-2) + 1(7)}{2+1} \right) = \left( \frac{6-3}{3}, \frac{2-2}{3}, \frac{-4+7}{3} \right) = (1, 0, 1)$. The direction ratios (DRs) of the line segment $AB$ are $(3 - (-3), 1 - (-2), -2 - 7) = (6, 3, -9)$. Since the plane is perpendicular to $AB$,the normal vector to the plane is $\vec{n} = (6, 3, -9)$,which can be simplified to $(2, 1, -3)$. The equation of the plane passing through $Q(1, 0, 1)$ with normal vector $(2, 1, -3)$ is: $2(x-1) + 1(y-0) - 3(z-1) = 0$ $2x - 2 + y - 3z + 3 = 0$ $2x + y - 3z + 1 = 0$ $2x + y - 3z = -1$ Dividing by $-1$: $-2x - y + 3z = 1$ $\frac{x}{-1/2} + \frac{y}{-1} + \frac{z}{1/3} = 1$. Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,the $y$-intercept is $b = -1$.

Let $A=(-3,-2,7)$ and $B=(3,1,-2)$. Let a plane perpendicular to the line segment $AB$ divide $AB$ in the ratio $2:1$. Then the intercept made by the plane on the $y$-axis is

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