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(C) The equation of a plane passing through the point $(2,3,1)$ is given by $a(x-2)+b(y-3)+c(z-1)=0$ $\ldots(1)$.Since the point $(4,-5,3)$ lies on th

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$x-z=1$

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(C) The equation of a plane passing through the point $(2,3,1)$ is given by $a(x-2)+b(y-3)+c(z-1)=0$ $\ldots(1)$.Since the point $(4,-5,3)$ lies on the plane,we substitute these coordinates into equation $(1)$:$a(4-2)+b(-5-3)+c(3-1)=0$$2a-8b+2c=0$$a-4b+c=0$ $\ldots(2)$.Since the plane is parallel to the $y$-axis,its normal vector is perpendicular to the $y$-axis vector $(0,1,0)$. Thus,$a(0)+b(1)+c(0)=0$,which implies $b=0$.Substituting $b=0$ into equation $(2)$,we get $a+c=0$,or $a=-c$.Substituting $a=-c$ and $b=0$ into equation $(1)$:$-c(x-2)+0(y-3)+c(z-1)=0$Dividing by $-c$ (assuming $c 
eq 0$):$(x-2)-(z-1)=0$$x-z-1=0$$x-z=1$.

The equation of the plane passing through the points $(2,3,1)$ and $(4,-5,3)$ and parallel to the $y$-axis is:

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