An equation of the plane containing the line x-1-3-y-3-2-z-2-1 and the point 0-7-7 is

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Equation of a Plane : Point Normal Form

An equation o...

Question

An equation of the plane containing the line $\frac{x + 1}{3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point (0, 7, -7) is

x + y + z = 0

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x + 2y - 3z = 35

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3x - 2y + 3z + 35 = 0

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3x - 2y - z = 21

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Solution

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\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}

(0, 7, - 7)

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}

(0, 7, - 7)

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}

\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

(1, 2, - 1)

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}

(0, 7, - 7),

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