The equation of the plane passing through the points $(2,3,1),(4,-5,3)$ and parallel to $\mathrm{y}$ -axis is
The equation of the plane passing through the points $(2,3,1),(4,-5,3)$ and parallel
to $\mathrm{y}$ -axis is
$x+z=3$
$x+z=1$
$x-z=1$
$z-x+2=0$
Solution
Equation of plane passing through the point $(2,3,1)$ is
$a(x-2)+b(y-3)+c(z-1)=0$ $\ldots(1)$
Given point $(4,-5,3)$ lies on plane
$2 a-8 b+2 c=0$ $\ldots(2)$
Since plane is parallel to $\mathrm{Y}$-axis, having d.r. $(0,1,0)$
(a) $(0)+(\mathrm{b})(1)+(\mathrm{c})(0)=0 \Rightarrow \mathrm{b}=0$
Putting in equation (2) we get
$2 a+2 c=0 \Rightarrow a=-c$
Putting values of $\mathrm{a}, \mathrm{b}$ in equation (1)
$\begin{aligned}
&-c(x-2)+c(z-1)=0 \\
\therefore &(x-2)-(z-1)=0 \\
\therefore & x-z=1
\end{aligned}$