## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 12 - Section 12.5 - Equations of Lines and Planes - 12.5 Exercises - Page 833: 71

#### Answer

$\frac{18}{7}$

#### Work Step by Step

Given: $(1,-2,4)$ ; $3x+2y+6z=5$ Distance formula between the point and the plane is given as: $D=\frac{|ax+by+cz+d|}{\sqrt {a^2+b^2+c^2}}$ $D=\frac{|3.1+2 \cdot (-2)+6 \cdot (4)-5|}{\sqrt {3^2+2^2+6^2}}$ $=\frac{18}{7}$

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