Given, $$ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$$
As we know, $$\left[ x \right] + \left\{ x \right\} = x$$
where $$\left[ x \right]$$ is integral part and $$\left\{ x \right\}$$ is fractional part.
$$\therefore$$$$\left\{ x \right\} = x - \left[ x \right]$$
Now put $$\left\{ x \right\}$$ inplace of $$x - \left[ x \right]$$ in the equation.
The new equation is $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} + {a^2} = 0$$

[Note : Question says this equation has no integral solution, it means $$\left\{ x \right\} \ne $$ 0. So, $$x$$ is not a integer.]

$$\therefore$$ $$\left\{ x \right\}$$ = $${{ - 2 \pm \sqrt {4 - 4 \times \left( { - 3} \right){a^2}} } \over { - 6}}$$
             = $${{ - 2 \pm \sqrt {4 + 12{a^2}} } \over { - 6}}$$
As $$\left\{ x \right\}$$ is fractional part so it is lies between 0 to 1($$0 \le \left\{ x \right\} < 1$$).

By considering positive sign, we get

$$0 \le {{ - 2 + \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 + \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge + \sqrt {4 + 12{a^2}} > - 4$$

$$\because$$$$ + \sqrt {4 + 12{a^2}} $$ is always positive which is greater than any negative no. So can ignore the inequality $$ + \sqrt {4 + 12{a^2}} > - 4$$

Consider this inequality,
$$2 \ge + \sqrt {4 + 12{a^2}} $$
$$ \Rightarrow $$ $$4 \ge 4 + 12{a^2}$$
$$ \Rightarrow $$ $$12{a^2} \le 0$$
$$ \Rightarrow $$ $${a^2} \le 0$$
$$ \Rightarrow $$ $${a^2} = 0$$
$$ \Rightarrow $$ $${a} = 0$$

If $$a$$ = 0 then $$ - 3{\left\{ x \right\}^2} + 2\left\{ x \right\} = 0$$ so $$\left\{ x \right\}$$ becomes 0 but question says $$\left\{ x \right\}$$ $$ \ne $$ 0.
So $$a$$ can't be 0.

Now by considering negative sign, we get


$$0 \le {{ - 2 - \sqrt {4 + 12{a^2}} } \over { - 6}} < 1$$

$$ \Rightarrow $$$$0 \ge - 2 - \sqrt {4 + 12{a^2}} > - 6$$

$$ \Rightarrow $$$$2 \ge - \sqrt {4 + 12{a^2}} > - 4$$

As 2 is always greater than $${ - \sqrt {4 + 12{a^2}} }$$. Ignore this inequality.

Now consider this inequality,

$$ - \sqrt {4 + 12{a^2}} > - 4$$

$$ \Rightarrow $$ $$\sqrt {4 + 12{a^2}} < 4$$

$$ \Rightarrow $$ $$4 + 12{a^2} < 16$$

$$ \Rightarrow $$ $$12{a^2} < 12$$

$$ \Rightarrow $$ $${a^2} < 1$$

$$ \Rightarrow $$ $$\left( {{a^2} - 1} \right) < 0$$

$$ \Rightarrow $$ $$\left( {a + 1} \right)\left( {a - 1} \right) < 0$$

$$ \Rightarrow $$ $$ - 1 < a < 1$$

But earlier we found that $$a$$ $$ \ne $$ 0.

So, the range of $$a$$ is = $$\left( { - 1,0} \right) \cup \left( {0,1} \right)$$