Find the equation of the plane passing through the points $(1, 0, - 1)$ and $(3, 2, 2)$ and parallel to the line $x - 1 = \frac{1 - y}{2} = \frac{z - 2}{3}$ .

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Solution

Let the plane equation be $a x + b y + c = d$ ;
$a - c = d; 3 a + 2 b + 2 c = d$
$a + 2 b + 3 c = 0 ⟹ a = 0, 2 b = 3 d, c = - d ⟹ 3 y - 2 z = 2$

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Similar questions

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

(3, - 1, 2)

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

\frac{x - 4}{3} = \frac{y}{- 1} = \frac{z}{2}

(4, - 1, 2)

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

\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}

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

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