Finding A point $Q$ on a surface such that Line $PQ$ get tangent to that surface 
Consider the surface $f: x^2 + y^2 - 2z^2 = 1$ and Point $P= (1,1,1)$.
  We want to find all points $Q$ on the surface such that line $PQ$ get
  tangent to the surface. Also we want to find the point $Q_0$ with
  above conditions such that $PQ_0$ gets smallest.

For the first part; I solved the problem like this: $\nabla f = (2x,2y,-4z) , Q=(x,y,z) \Rightarrow PQ : (x-1 , y-1 , z-1)$ and $PQ \perp \nabla f$ by using the dot product we get at last $2x^2 + 2y^2 -4z^2 -2x-2y +4 =0$ and using the equation of surface we get: $x+y-2z =1$ and $x^2 + y^2 -2z^2 = 1$.(Which is a curve - probably and elliptic curve) 
After these, I don't know how to find Q such that PQ gets smallest. I know that I must make $(x-1)^2 + (y-1)^2 + (z-1)^2$ smallest but I don't know how to use equations I got to do this.

Given point $P=(1,1,1)$, find set $S$ of all $Q=(x,y,z)$ on the surface $x^2+y^2-2z^2=1$ 
subject to the constraint that the line $PQ$ is tangent to the surface at $Q$. Then locate the $Q$ in that set closest to $P$.
OP has already found that the set $S$ satisfies the equations
$$x+y-2z =1 $$
$$x^2 + y^2 -2z^2 = 1 $$
So it remains to minimize
$$ R(x,y,z)=(x-1)^2+(y-1)^2+(z-1)^2 $$
subject to constraints (1) and (2).
Using the method of Lagrange multipliers, let 
$$f(x,y,z)=x^2+y^2-2z^2-1 \tag{1}$$
$$g(x,y,z)=x+y-2z-1 \tag{2}$$
Then solve
$$ \begin{cases}
\nabla R(x,y,z)-\lambda\nabla f(x,y,z)-\mu\nabla g(x,y,z)=0&\\
x+y-2z =1&\\
x^2 + y^2 -2z^2 = 1
\end{cases}$$
The gradient equation leads to
$$ x=y=\frac{\mu+2}{2(1-\lambda)} $$
which reduces the other two equations to
$$ \begin{cases}
x=z+\frac{1}{2}&\\x^2=z^2+\frac{1}{2}
\end{cases}  $$
giving $x=y=\frac{3}{4},\,z=\frac{1}{4}$. So the nearest $Q$ to $P$ is $Q=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$.
