If the system of linear equations

x + ky + 3z = 0
3x + ky - 2z = 0
2x + 4y - 3z = 0

has a non-zero solution (x, y, z), then $${{xz} \over {{y^2}}}$$ is equal to

If $$\left| {\matrix{ {x - 4} & {2x} & {2x} \cr {2x} & {x - 4} & {2x} \cr {2x} & {2x} & {x - 4} \cr } } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$

then the ordered pair (A, B) is equal to :

Two sets A and B are as under :

A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};

B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)² + 9(b - 5)² $$ \le $$ 36 };

Then

Let $${a_1}$$, $${a_2}$$, $${a_3}$$, ......... ,$${a_{49}}$$ be in A.P. such that

$$\sum\limits_{k = 0}^{12} {{a_{4k + 1}}} = 416$$ and $${a_9} + {a_{43}} = 66$$.

$$a_1^2 + a_2^2 + ....... + a_{17}^2 = 140m$$, then m is equal to