Correct Answer: $(2 \sqrt{17}-2)$

Solution :
In $ \triangle  $APQ and $ \triangle  $ABC,
$ \angle  $P = $ \angle  $B (corresponding angle)
$ \angle  $Q = $ \angle  $C (corresponding angle)
$ \triangle  $APQ $\sim$ $ \triangle  $ABC, by AA similarity.
Given that AP = QC, AB = 16 cm, and AQ = 4 cm.
Let the length of QC be $y$ cm.
We can use the property of similar triangles to set up the following proportion,
$⇒\frac{AP}{AB} = \frac{AQ}{AC}$
$⇒\frac{AB}{AP}= \frac{AC}{AQ}$
$⇒\frac{AB}{AP}-1= \frac{AC}{AQ}-1$
$⇒\frac{AB}{AP}-1= \frac{AC-AQ}{AQ}$
$⇒\frac{AB}{AP}-1= \frac{QC}{AQ}$
$⇒\frac{12}{QC}-1= \frac{QC}{4}$
$⇒\frac{12}{y}-1= \frac{y}{4}$
$⇒y^2+4y-64=0$
$⇒y=\frac{-4 \pm\sqrt{4^2-4 \times 1\times (-64)}}{2\times 1}$
$⇒y=\frac{-4 \pm\sqrt{272}}{2}$
$⇒y=\frac{-4 \pm 4\sqrt{17}}{2}$
$⇒y=(2 \sqrt{17}-2)$
Hence, the correct answer is $(2 \sqrt{17}-2)$.