# Chapter 13 - Vector Geometry - 13.5 Planes in 3-Space - Exercises - Page 685: 53

$$3\lambda x+by+2 \lambda z=5\lambda.\\ \lambda\neq 0.$$

#### Work Step by Step

The equation of any plane that intersects the $xz$-plane in the line $3x+2z=5$ is given by $$ax+by+cz=d.$$ Now, to find the $xz$ intersection, we put $y=0$ and hence we get $$ax+ cz=d.$$ Due to the intersection, we have $$a=3\lambda, \quad c=2\lambda, \quad d=5\lambda, \quad \lambda\neq 0.$$ So the equation of any plane whose intersection is the line $3x+2z=5$ is given by $$3\lambda x+by+2 \lambda z=5\lambda.$$

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