Answer
$\frac{7}{\sqrt{14}}$
Work Step by Step
The distance of a point $(a,b,c)$ from a plane $Ax+By+Cz=0$ is given by: $d=\frac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}}$, hence here: $d=\frac{|2(3)+3(2)-5|}{\sqrt{2^2+3^2+(-1)^2}}=\frac{7}{\sqrt{14}}$
Linear Algebra: A Modern Introduction
by Poole, David
Published by
Cengage Learning
ISBN 10:
1285463242
ISBN 13:
978-1-28546-324-7
Chapter 1 - Vectors - Chapter Review - Review Questions - Page 56: 14
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