## 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: 55

#### Answer

$(1,1,0)$

#### Work Step by Step

Since, we are given that the parametric equations $x=1+2t; y=1+5t; z=3t$ Equation of plane $x+y+z=2$ Plug all the parametric equations $x=1+2t; y=1+5t; z=3t$ in equation of plane, we have $t=0$ Now the Parametric equations are $x=1+2(0)=1; y=1+5(0)=1; z=3(0)=0$ Hence, the line will meet at the point: $(1,1,0)$

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