The full model is given by $E(Y_i\mid x)=\beta_0+\beta_1 x_{i1}+\beta_2 x_{i2} + \beta_3 x_{i3}$
Test the hypothesis that $E(Y_i\mid x)=x_{i1}$ is a good model.
I calculated the FULL Residual SS to be $RSS_{f}=0.765$
The Simple Residual SS is $RSS_s=1689.21$
Now I was to test that $\beta_0=\beta_2=\beta_3=0$ and $\beta_1=1$
My notes suggest to put it in the form $C\beta=d$ where $C$ is $c\times 4$ matrix.
Then I can do an F-test: $$f=\frac{(RSS_s-RSS_f)/c}{RSS_f/(n-4)}$$
Can somebody help me with building the $C\beta=d$ part? What is the value of $c$? Intuitively I think it's $c=1$
$c$ should be the number of restrictions or equivalently the difference between the "degrees of freedom" between the unrestricted and restricted model. In your case, $c=4-0=4$ (you have "restrictions" on all the $4$ regression coefficients). Formally,
$$ F=\frac{(C\hat{\beta} - d)'(C(X'X)^{-1}C')^{-1}(C\hat{\beta} - d)/c}{e'e/(n-p)}. $$ For your case you have $C=I$ and $d=(0,1,0,0)'$, thus $c=4$ and after some algebra you get the form of $$ F = \frac{(RSS_r - RSS_f)/4}{RSS_f/(n-4)}. $$