Answer
a) $2x+y-3z=27$
b) $x=\dfrac{27}{2}$
$y=27$
$z=-9$
Work Step by Step
a) The plane containing the point $P(-6,0,-3)$ and having the normal vector $n=\lt \dfrac{-2}{3}, \dfrac{-1}{3}, 1 \gt$ is expressed algebraically by the equation as follows:
$ \dfrac{-2}{3}(x-(-6)) -\dfrac{1}{3}(y-0)+1(z-(-3))=0$
or, $-2x-12-y+32+39=0 \implies 2x+y-3z=27$
b) To find the $x$-intercept, let us consider $y=z=0$
Thus, $2x=27 \implies x=\dfrac{27}{2}$
To find the $y$-intercept, let us consider $x=z=0$
Thus, $y=27$
To find the $z$-intercept, let us consider $x=y=0$
Thus, $-3z=27 \implies z=-9$
