$a, b, a\times b$ form a basis for $\Re^3$ hence $ r = ua+vb+w(a\times b)$ for scalars $u,v,w$
Sub in given equations to get
$ 0 + v(b\times a) + w( (a\cdotp a) b - (a\cdotp b)a = b $
$ u(a\cdotp c) + v(b\cdotp c) + w((a\times b) \cdotp c ) = \alpha $
Using $a\cdotp b = 0 $ the first equation gives v=0 and $ w = \cfrac{1}{||a||^2} $
The second equation is then $ u(a\cdotp c) = \alpha - \cfrac {(a\times b) \cdotp c}{||a||^2} $
If $a\cdotp c \ne 0 $ then $ r = \cfrac {\left(\alpha - \cfrac {(a\times b) \cdotp c}{||a||^2} \right)}{c\cdotp a} a + \cfrac {a\times b}{||a||^2}$
If $a\cdotp c=0$ and $\alpha \ne \cfrac {(a\times b) \cdotp c}{||a||^2} $ then there is no solution for $r$
If $a\cdotp c=0$ and $\alpha = \cfrac {(a\times b) \cdotp c}{||a||^2} $ then u is arbitrary and $r = ua +\cfrac {a\times b}{||a||^2}$
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This is the intersection of a plane with normal $c$ and a line with direction $a$.
If $a\cdotp c\ne 0$ they meet at a unique point.
If $a\cdotp c =0$ then they do not meet unless the line is entirely contained in the plane.
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