The distance between the planes $2 x - 3 y + 6 z + 12 = 0$ and $2 x - 3 y + 6 z - 2 = 0$ is

\frac{10}{7}

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\frac{2}{7}

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2

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\frac{24}{7}

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Solution

The correct option is C $2$

The given equations of the plane are $2 x - 3 y + 6 z + 12 = 0$ and $2 x - 3 y + 6 z + - 2 = 0$ .

The distance between the planes is $\frac{| d_{1} - d_{2} |}{\sqrt{a^{2} + b^{2} + c^{2}}}$

$= \frac{| 12 - (- 2) |}{\sqrt{2^{2} + 3^{2} + 6^{2}}}$

$=$ $\frac{14}{7}$

$=$ $2$

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The distance between the planes 2x−3y+6z+12=0 and 2x−3y+6z−2=0 is

The correct option is C 2 The given equations of the plane are 2x−3y+6z+12=0 and 2x−3y+6z+−2=0. The distance between the planes is |d1−d2|√a2+b2+c2 =|12−(−2)|√22+32+62 = 147 = 2

The distance between the planes $2 x - 3 y + 6 z + 12 = 0$ and $2 x - 3 y + 6 z - 2 = 0$ is