Find Point of intersection of the tangent plane to surface Find the point of intersection of the tangent plane to the surface $z+1=xe^y\cos(z)$ at the point $(1,0,0)$ and the line $L$ given by: $x=2t, y=t+1, z=1-3t$.

To specify the tangent plane we need a point on it (given) and its normal. This time the surface is given implicitly in the form $F(x,y,z)=0$ with 
$$
F(x,y,z)=x e^y\cos z-z-1.
$$
The normal is then given by the gradient
$$
\nabla F=(e^y\cos z, e^y x \cos z,-1-e^yx\sin z),
$$
which evaluated at $P=(1,0,0)$ gives us the normal vector
$$
\vec{n}=\nabla F(P)=(1,1,-1).
$$
Thus the equation of the tangent $T$ is of the form $x+y-z=a$. The constant $a$ can be solved using the fact that $P\in T$. We get $a=1$. Substituting the parametrizations into the equation of the tangent gives us the equation
$$
(2t)+(1+t)-(1-3t)=1.
$$
From which we solve $t=1/6$. The sought after point of intersection is thus
$$Q=(2\cdot\dfrac16,1+\dfrac16,1-3\cdot\dfrac16)=(1/3,7/6,1/2).$$
Here's a Mathematica pic of the surface (the blue thingy with mesh lines), the tangent plane (the semi-opaque grey plane), the line (red) and the point $Q$.

