Given the curve r(a)=<6−a^2,a^3+1,1−a> find all the points where the tangent vector on r(a)

The question is the following:
Given the curve $r\left(a\right)=<6-{a}^{2},{a}^{3}+1,1-a>$ and the plane $x+y+z=\pi$, find all the points where the tangent vector on r(a) is parallel to the plane.
I know finding the tangent vector is the first part of the problem. That would be $T\left(a\right)=\frac{<-2a,3{a}^{2},-1>}{\sqrt{\left(-2a{\right)}^{2}+\left(3{a}^{2}{\right)}^{2}+\left(-1{\right)}^{2}}}$. But beyond there I don't know how to draw a relationship between the line and plane.
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luthersavage6lm
A vector $\left(\alpha ,\beta ,\gamma \right)$ is parallel to that plane if and only if $\alpha +\beta +\gamma =0$. Therefore,