Question:
Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
A(8, -1, -19) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to ___________.
Updated On: Mar 4, 2024
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Correct Answer: 26
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The answer is: 26
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