Question

Let \( P(x_1, y_1, z_1) \) be the foot of the perpendicular drawn from the point \[ Q(2, -2, 1) \] to the plane \[ x - 2y + z = 1. \] If \( d \) is the perpendicular distance from the point \( Q \) to the plane and \[ I = x_1 + y_1 + z_1, \] then \( I + 3d^2 \) is:

• A. \( 5 \)
• B. \( 7 \)
• C. \( 19 \)
• D. \( 26 \)

Hint:

The perpendicular distance formula is useful for calculating distances from a point to a plane in 3D geometry.

Answer 1

The Correct Option is C

Step 1: Finding the perpendicular distance \( d \) The perpendicular distance from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz + d = 0 \) is given by: \[ d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] Substituting the given values: \[ d = \frac{|(1)(2) + (-2)(-2) + (1)(1) - 1|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|2 + 4 + 1 - 1|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}. \] Step 2: Computing \( I + 3d^2 \) Given \( I = x_1 + y_1 + z_1 \), solving using the foot of the perpendicular formula, we find: \[ I + 3d^2 = 19. \]