Given data

$|\vec a|=1,\quad |\vec b|=4,\quad \vec a\cdot \vec b=2$

Also, $\vec c=2(\vec a\times \vec b)-3\vec b$

We need to find $192\sin^2\alpha$ where $\alpha$ is the angle between $\vec b$ and $\vec c$.

Find $|\vec a\times \vec b|$

Using $|\vec a\times \vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot \vec b)^2$

we get $|\vec a\times \vec b|^2=(1^2)(4^2)-2^2=16-4=12$

Hence, $|\vec a\times \vec b|=\sqrt{12}=2\sqrt 3$

Use perpendicularity

Since $\vec a\times \vec b$ is perpendicular to $\vec b$, $\vec b\cdot (\vec a\times \vec b)=0$

Therefore, in $\vec c=2(\vec a\times \vec b)-3\vec b,$ the two components are perpendicular.

Find $\vec b\cdot \vec c$

$\vec b\cdot \vec c=\vec b\cdot\left(2(\vec a\times \vec b)-3\vec b\right)$ $=2\vec b\cdot(\vec a\times \vec b)-3|\vec b|^2$ $=0-3(16)=-48$

Find $|\vec c|^2$

Because the two parts are perpendicular, $|\vec c|^2=|2(\vec a\times \vec b)|^2+|-3\vec b|^2$ $=4|\vec a\times \vec b|^2+9|\vec b|^2$ $=4(12)+9(16)=48+144=192$

So, $|\vec c|=\sqrt{192}=8\sqrt 3$

Find $\cos\alpha$

$\cos\alpha=\frac{\vec b\cdot \vec c}{|\vec b||\vec c|}$ $=\frac{-48}{4\cdot 8\sqrt 3}$ $=\frac{-48}{32\sqrt 3}=-\frac{\sqrt 3}{2}$

Find $\sin^2\alpha$

$\sin^2\alpha=1-\cos^2\alpha=1-\left(\frac{\sqrt 3}{2}\right)^2=1-\frac34=\frac14$

Compute required value

$192\sin^2\alpha=192\cdot \frac14=48$

Comparison with stored answer

Our derived answer is 48, which matches the stored correct answer.