Answer
a) $6x-z-4=0$
b) $x-intercept =\dfrac{2}{3}$
$ z-intercept =-4$
Work Step by Step
a) The plane containing the point $P(2,4,8)$ and having the normal vector $n=\lt 3,0,\dfrac{-1}{2} \gt$ is expressed algebraically by the equation as follows:
$3(x-2)+0(y-4)+(\dfrac{-1}{2})(z-8)=0$
or, $3x-6-\dfrac{z}{2}+4=0 \implies 6x-z-4=0$
b) To find the $x$-intercept, let us consider $z=0$
Thus, $6x-4=0 \implies x=\dfrac{2}{3}$
To find the $z$-intercept, let us consider $x=0$
Thus, $-z-4=0 \implies z=-4$ and $y$-intercept $=\infty$
