Prove that surfaces $x + 2y – lnz + 4 = 0$ and $x^2 - xy – 8x + z + 5 = 0$ are tangent at $(2,-3,1)$. $x + 2y – ln(z) + 4 = 0$
$x^2 - xy – 8x + z + 5 = 0$
$\nabla [1,2,\frac{-1}{z}] $
$\nabla [2x-y-8,-x,1] $
$\nabla(P0) [1,2,-1] $
$\nabla(P0) [-1,-2,1] $
I've stuck at this point and i don't know what to do next.
 A: Two surfaces $S_1$ and $S_2$ are tangent at a point $P$ if and only if 
$P \in S_1 \cap S_2, \tag{1}$
i.e., $P$ lies in each of $S_1$, $S_2$; and 
$T_PS_1 = T_PS_2, \tag{2}$
that is, the tanget planes to $S_1$ and $S_2$ at $P$ are the same.  Taking $S_1$ to be the surface
$f_1(x, y, z) = x + 2y - \ln z + 4 = 0, \tag{3}$
and $S_2$ to be
$f_2(x, y, z) = x^2 - xy - 8x + z + 5 = 0, \tag{4}$
we first verify
$P = (2, -3, 1) \in S_1 \cap S_2 \tag{5}$
by showing
$f_1(P) = f_2(P) = 0; \tag{6}$
we have
$f_1(2, -3, 1)= 2 + 2(-3) -\ln 1 + 4 = 2 - 6 + 4 = 0, \tag{7}$
and
$f_2(2, -3, 1) = 2^2 - 2(-3) - 8(2) + 1 + 5 =  4 + 6 - 16 + 1 + 5= 0; \tag{8}$
thus
$P \in S_1 \cap S_2. \tag{9}$
To see that the tangent planes to $S_1$ and $S_2$ at $P$ coincide, we may show the normal vectors to each surface at $P$ are collinear.  We do this by calculating $\nabla f_1$ and $\nabla f_2$ at $P$, viz.
$\nabla f_1 = (1, 2, -\dfrac{1}{z}), \tag{10}$
$\nabla f_2 = (2x - y - 8, -x, 1); \tag{11}$
thus
$\nabla f_1(P) = (1, 2, -1) \tag{12}$
and
$\nabla f_2(P) = (2(2) - (-3) - 8, -2, 1) = (-1, -2, 1); \tag{13}$
we see that
$\nabla f_2(P) = -\nabla f_1(P).  \tag{14}$
Since the normals to $S_1$ and $S_2$ are collinear at the point of intersection $P$, their tangent planes are one and the same there; the surfaces are indeed tangent at $P$.
