## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 23

#### Answer

$7x-5y-4z=6$

#### Work Step by Step

The plane passing through the points $(1,1,-1),(2,0,2), (0,-2,1)$ Now, $\overrightarrow{a}=\lt 1,-1,3 \gt; \overrightarrow{b}=\lt -1,-3,2 \gt$ and $\overrightarrow{a}\times \overrightarrow{b}=\lt -1,-3,2 \gt= 7i-5j-4k$ Suppose the equation of plane is $7x-5y-4z=D$ For point $(1,1,-1)$: $7(1)-5(1)-4(-1)=d \implies d=6$ Hence, $7x-5y-4z=6$

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