Answer
(a) $2x-2y-3z=3$
(b) $\frac{x-2}{4}=\frac{y+1}{-4}=\frac{z-1}{-6}$
Work Step by Step
$x^2+2y^2-3z^2=3$ and $(2,-1,1)$
(a) Equation for a tangent plane is given as:
$4(x-2)+(-4)(y+1)+(-6)(z-1)=0$
$4x-4y-6z=6$
$2x-2y-3z=3$
(b) $2x-2y-3z=3$
Equation of normal line is:
$\frac{x-2}{4}=\frac{y+1}{-4}=\frac{z-1}{-6}$
