Finding minimum of value $x^2+y^2+z^2+t^2$ 
Suppose $x,y,z,t$ are real numbers which holds for these equations:
numbers:
${ \begin{cases}{x^2+6y+2t=2} \\ {y^2+2t+y=-14} \\
 {z^2+4x+2y=-27} \\  {t^2+2z+y=5}\end{cases} }$
What is the minimum
value of $x^2+y^2+z^2+t^2$ ?
$$1)12\quad\quad\quad\quad\quad\quad2)18\quad\quad\quad\quad\quad\quad3)34\quad\quad\quad\quad\quad\quad4)38\quad\quad\quad\quad\quad\quad5)\text{none}$$
$$$$

I added all the equations together:
$$x^2+y^2+z^2+t^2+4x+10y+4t+2z=-34$$
$$(x^2+4x+4)+(y^2+10y+25)+(z^2+2z+1)+(t^2+4t+4)=0$$
$$(x+2)^2+(y+5)^2+(z+1)^2+(t+2)^2=0$$
Since each squares are non negative, each of them should be zero, therefore $x=-2,\quad y=-5,\quad z=-1,\quad t=-2.$
But when I put these numbers in the original equations neither of them holds for the equations. so can we  conclude there is not exist such real numbers $x,y,z,t$ that holds for the equations and therefore the answer is $5)\text{none}$. am I right?

I have both checked your solution, and confirmed it via WolframAlpha. No real solutions exist for this system of equations, and hence, the answer should be none.
Here is the link to the WolframAlpha computation.
