Answer
$\omega_{BC}=2.45 rad/s$
$v_C=2.20ft/s$
Work Step by Step
The required velocity and angular velocity can be determined as follows:
We know that
$\vec{v_C}=\vec{v_B}+\vec{\omega_{BC}}\times \vec{r_{C/B}}~$[eq(1)]
Similarly $\vec{v_C}=-\vec{v_C}\hat i$
and $\vec{r_{C/B}}=3cos45\hat i+3sin45\hat j=2.1213\hat i+2.1213\hat j$
The velocity of $B$ is given as
$\vec{\omega_B}=\vec{\omega_{AB}}\times \vec{r_{AB}}$
$\vec{\omega_B}=3\hat k\times (-2cos 30\hat i-2sin 30\hat j)=-5.1961\hat j
+3\hat i$
We plug in the known values in eq(1) to obtain:
$\vec{-v_C}\hat i=(-5.1961\hat j+3\hat i)+(\omega_{BC})\hat k\times (2.1213\hat i+2.1213 \hat j)$
$\implies \vec{-v_C}\hat i=(3-2.1213\omega_{BC})\hat i+(2.1213\omega_{BC}-5.1961)\hat j$
Comparing $i$ components, we obtain:
$-v_C=3-2.1213\omega_{BC}$
$\implies v_C=2.1213\omega_{BC}-3~$[eq(2)]
Now comparing $j$ component, we obtain:
$0=2.1213\omega_{BC}-5.1961$
This simplifies to:
$\omega_{BC}=2.45 rad/s$
We plug in this value in eq(2) to obtain:
$v_C=2.1213(2.449)-3$
$v_C=2.20ft/s$
