What will come at place of x, (x < 10) for $$\frac{(132 \div 12 \times x - 3 \times 3)}{(5^2 - 6 \times 4 + x^2)} = 1$$ ?

$$\frac{(132\div12\times x-3\times3)}{(5^2-6\times4+x^2)}=1$$

$$=$$>  $$\frac{(11\times x-3\times3)}{(25-6\times4+x^2)}=1$$

$$=$$>  $$\frac{11x-9}{25-24+x^2}=1$$

$$=$$>  $$\frac{11x-9}{1+x^2}=1$$

$$=$$>  $$11x-9=1+x^2$$

$$=$$>  $$x^2-11x+10=0$$

$$=$$>  $$x^2-10x-x+10=0$$

$$=$$>  $$x\left(x-10\right)-1\left(x-10\right)=0$$

$$=$$>  $$\left(x-10\right)\left(x-1\right)=0$$

$$=$$>  $$x=10$$  or  $$x=1$$

Given $$x < 10$$

$$=$$>  $$x=1$$

Hence, the correct answer is Option B