Remove a parallel component from an area calculation

$\sqrt{126}=\frac12|\mathbf a\times(\lambda\mathbf a+\mu\mathbf b)|=\frac12|\lambda(\mathbf a\times\mathbf a)+\mu(\mathbf a\times\mathbf b)|$. The self-cross-product is zero, so this is $\frac\mu2|\mathbf a\times\mathbf b|$. Here $\mathbf a\times\mathbf b=(-2,1,3)$, whose magnitude is $\sqrt{14}$. Thus $\sqrt{126}=\mu\sqrt{14}/2$ and μ=6, using μ>0.

$\mathbf c=(\lambda+\mu,-\lambda+2\mu,\lambda)$. Direct calculation gives $\mathbf a\times\mathbf c=(-2\mu,\mu,3\mu)=\mu(-2,1,3)$. Consequently $\sqrt{126}=\mu\sqrt{14}/2$ again, giving μ=6.

For μ=4, $\mathbf c=(\lambda+4,-\lambda+8,\lambda)$. Squaring $|\mathbf c|=5\sqrt3$ gives $(\lambda+4)^2+(-\lambda+8)^2+\lambda^2=75$, hence $3\lambda^2-8\lambda+5=(\lambda-1)(3\lambda-5)=0$. Both λ=1 and 5/3 are positive. Thus $C=(5,7,1)$ or $(17/3,19/3,5/3)$.