Concept:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d and a point whose coordinate is, (x1, y1, z1).

Then the distance between the point and the plane is given by

 \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Quadratic equation with roots α and β:

Quadratic equation with roots α and β is given by

(x - α)(x - β) = 0  

Calculation:

We know that the distance between the point and the plane is given by

\(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

So, the distance of the point (2, 3, -5) from the given plane

 \(α\ =\ \left| {\frac{{2\; \times \;1\; + \ {2}\ \times \;3\; + \;(-\ 2)\ \times \ (-\ 5)\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\ =\ 3\)

Similarly, the distance of the point (3, 1, 1) from the given plane 

\(β \ =\ \left| {\frac{{3\; \times \;1\; + \ {2}\ \times \;1\; + \;(-\ 2)\ \times \ 1\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\ =\ 2\)

Therefore, quadratic equation with roots 3 & 2 will be

(x - 2)(x - 3) = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x2 - 5x + 6 = 0

Hence, option 3 is correct.