Question

The distance of the point $(-1,9,-16)$ from the plane $2 x+3 y-z=5$ measured parallel to the line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ is

• A. $20 \sqrt{2}$
• B. 31
• C. $13 \sqrt{2}$
• D. 26

Answer 1

The Correct Option is D

The equation of the line is: \[ \frac{x + 4}{3} = \frac{y - 9}{-4} = \frac{z - 2}{12}. \] Using the parametric form of the line: \[ x = 3\lambda - 4, \quad y = -4\lambda + 9, \quad z = 12\lambda + 2. \] Substitute these into the plane equation \( 2x + 3y - z = 5 \): \[ 2(3\lambda - 4) + 3(-4\lambda + 9) - (12\lambda + 2) = 5. \] Simplify: \[ 6\lambda - 8 - 12\lambda + 27 - 12\lambda - 2 = 5, \] \[ -18\lambda + 17 = 5 \implies \lambda = \frac{2}{-18} = \frac{-2}{9}. \] The point of intersection is: \[ x = 3\left(\frac{-2}{9}\right) - 4 = -2 - 4 = -6, \] \[ y = -4\left(\frac{-2}{9}\right) + 9 = \frac{8}{9} + 9 = 10, \] \[ z = 12\left(\frac{-2}{9}\right) + 2 = -\frac{24}{9} + 2 = 2. \] The distance between \((-1, 9, -16)\) and \((5, 1, 8)\) is: \[ \text{Distance} = \sqrt{(5 + 1)^2 + (1 - 9)^2 + (8 - (-16))^2}, \] \[ = \sqrt{6^2 + 8^2 + 24^2} = \sqrt{36 + 64 + 576} = \sqrt{676} = 26. \] Thus, the distance is: \[ \boxed{26}. \]