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

The correct answer is Option (3) → 3

The distance of a point from a plane ax + by + cz = d is \(\displaystyle\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}\). Here

    \( \displaystyle \frac{|2 + 2(3) - 2(-5) - 9|}{\sqrt{1+4+4}} = \frac{|2+6+10-9|}{3} \)

    \( \displaystyle = \frac{9}{3} = 3 \)

Note: This topic was not in the CUET 2026 syllabus.