Projection on a Plane

Let, $\hat{a}=B\hat{b}+C\hat{c}+N\hat{n}~...~...~...(i)$ , where, $\hat{n}$ is the unit normal vector on the plane $P$ . Now, $\hat{a} \cdot \hat{b} =B \hat{b} \cdot \hat{b} + C \hat{c} \cdot \hat{b} +N \hat{n} \cdot \hat{b}$ $\rightarrow \cos{\theta_{ab}}=B+C \cos{\theta_{bc}}~...~ ... ~...(ii)$ Again, taking the dot product of $\hat{c}$ and eqn. (i), $\cos{\theta_{ac}}=C+B \cos{\theta_{bc}}~...~ ...~ ...(iii)$ Now, solving eqn. (ii) & (iii), $B=\frac{\cos{\theta_{ab}}-\cos{\theta_{ac}} \cos{\theta_{bc}}}{1-\cos^2{\theta_{bc}}}$ $C=\frac{\cos{\theta_{ac}}-\cos{\theta_{ab}} \cos{\theta_{bc}}}{1-\cos^2{\theta_{bc}}}$ Plugging in the values of $\cos{\theta_{ab}}, \cos{\theta_{ac}}$ & $\cos{\theta_{bc}}$ , we get, $B=\frac{7}{45} ;~~ C=\frac{4}{45}$ So, $B-C=\frac{1}{15}$ ,i.e, the answer is $16$