It depends on what you mean by solve. Dr Binder is of course correct, it is a quadratic surface rather than a curve. Notice that the quadratic curve y = f(x) with y in R1, maps R1 to R1 or, if we allow complex roots, C1 to R1. If we take your equation as implicitly defining a quadratic surface z = f(x,y), then we may think of solving it as finding an explicit formula mapping values (x,y) to z. If we allow for complex values, then there is always a solution, and the map is R2 to C1. But if we require z in Reals, there may or may not be a Real z for a given Real (x,y). In the case of your equation, there is a solution over the Reals for exactly one point in R2.

In[1]:= (* your equation in proper Mathematica form *)
eq = x^2 + 2 y^2 + 2 z^2 + 2 x y + 2 y z - 2 x - 6 y - 10 z + 14 == 0;

In[2]:= (* solution for z in terms of x and y *)
sol = Solve[eq, z]

Out[2]= {{z -> 
   1/2 (5 - y - Sqrt[-3 + 4 x - 2 x^2 + 2 y - 4 x y - 3 y^2])}, {z -> 
   1/2 (5 - y + Sqrt[-3 + 4 x - 2 x^2 + 2 y - 4 x y - 3 y^2])}}

In[3]:= (* separated *)
{z1, z2} = z /. sol

Out[3]= {1/2 (5 - y - Sqrt[-3 + 4 x - 2 x^2 + 2 y - 4 x y - 3 y^2]), 
 1/2 (5 - y + Sqrt[-3 + 4 x - 2 x^2 + 2 y - 4 x y - 3 y^2])}

In[4]:= (* over what domain in Reals can we expect Real solutons *)
Reduce[z1 \[Element] Reals, {x, y}, Reals]

Out[4]= x == 2 && y == -1

In[5]:= Reduce[z2 \[Element] Reals, {x, y}, Reals]

Out[5]= x == 2 && y == -1

In[6]:= (* so there is exactly one point where the radical expression \
is Real, and there it is zero *)
sol /. {x -> 2, y -> -1}

Out[6]= {{z -> 3}, {z -> 3}}