Question

Let the foot of the perpendicular from the point \( (1, 2, 4) \) on the line \( \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} \) be \( P \). Then the distance of \( P \) from the plane \( 3x + 4y + 12z + 23 = 0 \) is:

• A. \( 5 \)
• B. \( \frac{50}{13} \)
• C. \( 4 \)
• D. \( \frac{63}{13} \)

Hint:

To calculate the perpendicular distance from a point to a plane, use the formula: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]