The largest integral value of ‘a’ for which every solution of the equation,  $x,(\left[x\right]-5)+2\left\{x\right\}+6=0$ satisfies the inequality $(a-3)x^2+2(a+3)x-8a\le 0$ , where [y] and {y} denote greatest integer and fractional part function respectively.

$x\left[x\right]-5x+2x-2\left[x\right]+6=0$

$\Rightarrow \left[x\right](x-2)-3(x-2)=0$

$\Rightarrow (\left[x\right]-3)(x-2)=0\Rightarrow x=2or[x=3]$

$\Rightarrow x\in [3,4)\cup \left\{2\right\}$

 Now,  $(a-3)x^2+2(a+3)x-8a\le 0$

$\Rightarrow a{x}^2-3x^2+2ax+6x-8a\le 0$

$\Rightarrow a{x}^2-2ax-3x^2+6x+4ax-8a\le 0$

$\Rightarrow ax(x-2)-3x(x-2)+4a(x-2)\le 0$

$\Rightarrow (x-2)(x(a-3)+4a)\le 0$

Case-I  $a-3>0\Rightarrow a>3$

$\frac{-4a}{a-3}\ge 4\Rightarrow -4a\ge 4a-12$

$8a\le 12\Rightarrow a\le \frac{3}{2}$

No solution

Case-II   $a-3<0$

$\frac{-4a}{a-3}\le 2\Rightarrow -4a\ge 2a-6$