# Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 22

$3x+y+z=5$

#### Work Step by Step

When the plane passing through the point $P(a,b,c)$ and normal to vector $n=pi+qj+rk$ then we have a component equation as follows: $a(x-a)+b(y-b)+c(z-c)=0$ ...(x) Here, $P(1,-1,3)$ and $n=3i+j+k$ From Equation (x), we get $3(x-1)+1(y+1)+1(z-3)=0$ $\implies 3x-3+y+1+z-3=0$ Thus, $3x+y+z=5$

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