Answer
(a) $x+2y+5z=0$
(b) $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-0}{5}=t$
or
$x=t+2$, $y=2t-1$, $z=5t$
Work Step by Step
$sin(xyz)=x+2y+3z ; (2,-1,0)$
(a) Equation for a tangent plane is given as:
$z-0=-\frac{1}{5}(x-2)-\frac{2}{5}(y+1)$
$5z=-(x-2)-2(y+1)$
$x+2y+5z=0$
(b) $x+2y+5z=0$
Equation of normal line is:
$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-0}{5}=t$
or
$x=t+2$, $y=2t-1$, $z=5t$
