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The distance of the point (2,3,−5)(2, 3, -5) from the plane x+2y−2z=9x + 2y - 2z = 9 is :

Solution

✅ Correct Option: 3

Distance =∣x0+2y0−2z0−9∣12+22+(−2)2= \frac{|x_0 + 2y_0 - 2z_0 - 9|}{\sqrt{1^2 + 2^2 + (-2)^2}}

=∣2+6+10−9∣9= \frac{|2 + 6 + 10 - 9|}{\sqrt{9}}

=93=3= \frac{9}{3} = 3.

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