The equation x + 2y + 2z = 1 and 2x + 4y + 4z = 9 have

If x, y and z are real and different and $$\,u = {x^2} + 4{y^2} + 9{z^2} - 6yz - 3zx - 2xy$$, then u is always.

Given that $${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$$
where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$
Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$

$${}^n{C_{r - 1}} = 36,{}^n{C_r} = 84\,\,and\,\,{}^n{C_{r + 1}} = 126$$, then r is :