If  $\left|\begin{array}{ccc}{ }^9{C}_4& { }^9{C}_5& { }^{10}{C}_r\\ { }^{10}{C}_6& { }^{10}{C}_7& { }^{11}{C}_{r+2}\\ { }^{11}{C}_8& { }^{11}{C}_9& { }^{12}{C}_{r+4}\end{array}\right|=0$ , then the value of   $6(\frac{{ }^r{C}_1}{2}+\frac{{ }^r{C}_2}{3}+\frac{{ }^r{C}_3}{4}+\frac{{ }^r{C}_4}{5}+\frac{{ }^r{C}_5}{6})$

${ }^n{C}_{r-1}{+}^n{C}_r{=}^{(n+1)}{C}_r$ , $C_{2}N{C}_2+C_1$
$C_2~C_2+C_1\Rightarrow \left|\begin{array}{ccc}{ }^9{C}_4& { }^9{C}_5& { }^{10}{C}_r\\ { }^{10}{C}_6& { }^{10}{C}_7& { }^{11}{C}_{r+2}\\ { }^{11}{C}_8& { }^{11}{C}_9& { }^{12}{C}_{r+4}\end{array}\right|=0$  is possible,   $\iff C_2\cong C_3\iff r=5$