Finding the non-zero value of ω, such that V and I will be in phase
Try to write your final expression in the form \$a+jb\$ by multiplying numerator and denominator with complex conjugate of denominator. 

Then  equate b = 0 and solve for \$\omega\$.

EDIT

Multiplying numerator and denominator of 1st term with conjugate of denominator,
$$Z_{eq} =\frac{j120ω^2+6ω}{20ω+j\left(12ω^2-1\right)} \times \frac{20ω-j\left(12ω^2-1\right)}{20ω-j\left(12ω^2-1\right)} + 1$$
$$Z_{eq} =1+\frac{120ω^2+120ω^2\left(12ω^2-1\right)}{(20ω)^2+\left(12ω^2-1\right)^2} +j \frac{2400ω^3-6ω\left(12ω^2-1\right)}{(20ω)^2+\left(12ω^2-1\right)^2} $$
Equating imaginary part to zero,
$$\frac{2400ω^3-6ω\left(12ω^2-1\right)}{(20ω)^2+\left(12ω^2-1\right)^2} =0$$

Solving this will produce
\$\omega = 0,\omega = \pm j\sqrt{388},\omega=\infty\$ as the solutions.