In the case when a=ai, b=bj and c=ck, where a, b, and c are positive scalar constants, determine the equation of the plane (which contains a, b, and c) in the form r.n=d, where the components of n and the value of the scalar constant d are to be given in terms of a, b, and c.
I'm having trouble with this question because I don't know what to cross to find n. Can anyone help me solve this?
If the interpretation of your question is the way that Ross Millikan suggested, then you have to solve the problem as though there are three points in the plane $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$. The hint that Ross Millikan gave you was to find the two vectors in the plane. And its clear, from your comments, that you found the normal vector which if I am not mistaken is $$n=\begin{pmatrix} -a \\[0.3em] b \\[0.3em] 0 \\[0.3em] \end{pmatrix} \times\begin{pmatrix} -a \\[0.3em] 0 \\[0.3em] c \\[0.3em] \end{pmatrix}=\begin{pmatrix} bc \\[0.3em] ac \\[0.3em] ab \\[0.3em] \end{pmatrix}$$ You can find $d$ if you first find the Cartesian equation which is by expanding $n\cdot r=(bc )x+(ac)y+ab(z)$. You then replace a point, say $A$, and find that $d=abc$.
Therefore the normal vector equation of the plane is,
$$\begin{pmatrix} bc \\[0.3em] ac \\[0.3em] ab \\[0.3em] \end{pmatrix} \bullet \ r=abc$$