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Question
Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.
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Solution
\[ \text{ Let the equation of a plane parallel to the given plane be } \]
\[2x - 3y + 5z = k . . . \left( 1 \right)\]
\[\text{ This passes through (3, 4, -1).} \hspace{0.167em} \text{ So } ,\]
\[2 \left( 3 \right) - 3 \left( 4 \right) + 5 \left( - 1 \right) = k\]
\[ \Rightarrow k = -11 \]
\[\text{ Substituting this in (1), we get } \]
\[2x - 3y + 5z = -11 . . . \left( 1 \right), \text{ which is the equation of the required plane } .\]
\[\text{ The equation of the given plane is } \]
\[2x - 3y + 5z = - 7 . . . \left( 2 \right)\]
\[ \text{ We know that the distance between two planes ax + by }+ cz = d_{ 1} \text{ and } ax + by + cz = d_{2 } \text{ is } \frac{\left| d_{2 }- d_1 \right|}{\sqrt{a^2 + b^2 + c^2}}\]
\[\text{ So, the required distance } \]
\[ = \frac{\left| - 7 - \left( - 11 \right) \right|}{\sqrt{2^2 + \left( - 3 \right)^2 + 5^2}}\]
\[ = \frac{\left| - 7 + 11 \right|}{\sqrt{4 + 9 + 25}}\]
\[ = \frac{4}{\sqrt{38}} \text{ units } \]
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