Answer
$x+y-z=5$
$x=5$
$y=5$
$z=-5$
Work Step by Step
The plane containing the point $P(p,q,r)$ and having the normal vector $n=\lt a,b,c\gt$ is expressed algebraically by the equation as follows:
$a(x-p)+b(y-q)+c(z-r)=0$
or, $1(x-0)+1(y-2)+(-1)(z-(-3))=0$
or, $x+y-2-(z+3)=0 \implies x+y-z=5$
To find the $x$-intercept, let us consider $y=0; z=0$
Thus, $x=5$
To find the $y$-intercept, let us consider $x=0; z=0$
Thus, $y=5$
To find the $z$-intercept, let us consider $x=0;y=0$
Thus, $z=-5$
