Pairs of Quadratic equations If each pair of the three equations
 $$x^2 + P_1x + q_1 = 0$$
$$x^2 + P_2x + q_2 = 0$$
$$x^2 + P_3x + q_3 = 0$$ 
have a common root, then prove that 
$$P_1^2 + P_2^2 + P_3^2 + 4(q_1 + q_2 + q_3) = 2 ( P_1.P_2 + P_2.P_3 + P_3.P_1)$$

This statement is false.  Rewrite the first two equations in the form:
$\left[\begin{array}{rcl}x^2&x&1\end{array}\right]\left[\begin{array}{c}1\\P_1\\q_1\end{array}\right] = \left[\begin{array}{rcl}x^2&x&1\end{array}\right]\left[\begin{array}{c}1\\P_2\\q_2\end{array}\right]
%\left[\begin{array}{c}0\\0\\0\end{array}\right]$.
To solve this system of equations, we move $x$'s to one side and cancelling $x^2$'s, to get $x (P_1-P_2) = q_2-q_1$.  So for this solution to exist, it is necessary and sufficient that $P_1\neq P_2$ or $q_2 = q_1$.
This does not tell you anything about the relationship between the $P_i$ and $q_i$.  In particular, if we let all $q_i=0$, then all three pairs will have the solution $x=0$; however, we can choose $P_1=1=P_3$ and $P_2=-1$ to get 1+1+1+4(0) = 2(-1+1+-1) = -2, contradiction.
