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प्रश्न
Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.
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उत्तर
\[ \text{ The given equation of the plane is } \]
\[2x - 3y + 6z + 14 = 0\]
\[2x - 3y + 6z = - 14 . . . \left( 1 \right)\]
\[ \text{ Now } ,\sqrt{2^2 + \left( - 3 \right)^2 + \left( 6 \right)^2}=\sqrt{4 + 9 + 36}=\sqrt{49}= 7\]
\[ \text{ Dividing (1) by 7, we get } \]
\[\frac{2}{7}x - \frac{3}{7}y + \frac{6}{7}z = - 2 \]
\[ \text{ Multiplying both sides by -1, we get } \]
\[ - \frac{2}{7}x + \frac{3}{7}y - \frac{6}{7}z = 2\]
\[ \text{ This is the normal form of the given equation of the plane } .\]
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