The intersection of $5$ planes If the intersection of planes $a^{k}x+2^{k}y+3^{k}z+d^{k}=0(0\leq k\leq 4)$ is a line, then $a=?$ and $d=?$.
 A: If you rewrite it as $a^kx + 2^ky + 3^k z + d^kw=0$ then it's the intersection of the null-space of the matrix $\begin{pmatrix}a^k & 2^k & 3^k & d^k\end{pmatrix}$ and the hyper-plane $w=1$. This being a line would require the null-space to be a plane - which means that the rank of the matrix is $2$. 
Since the second and third columns are independent this means that the first and last must be a linear combination of those which is only possible if $a$ and $d$ are $2$ or $3$.
A: The matrix of coefficients is 
$$A=\left(
\begin{array}{cccc}
 1 & 1 & 1 & 1 \\
 a & 2 & 3 & d \\
 a^2 & 4 & 9 & d^2 \\
 a^3 & 8 & 27 & d^3 \\
 a^4 & 16 & 81 & d^4 \\
\end{array}
\right)$$
The system has One and only one solution, that is one point, if $\text{rank }A=4$
We know that the planes intersect along a line, and this means that $\text{rank }A=3$ which happens if determinants of order $4$ are all zero.
$$\det \left|
\begin{array}{cccc}
 1 & 1 & 1 & 1 \\
 a & 2 & 3 & d \\
 a^2 & 4 & 9 & d^2 \\
 a^3 & 8 & 27 & d^3 \\
\end{array}
\right|=\\=-a^3d^2+5 a^3 d-6 a^3+a^2 d^3-19 a^2 d+30 a^2-5 a d^3+19 a d^2-36 a+6 d^3-30 d^2+36 d$$
can be easily factored as $$(a-3) (a-2) (d-3) (d-2) (d-a)=0$$
$(a=2,\;d=2);\;(a=2,\;d=3);\;(a=3,\;d=2);\;(a=3,\;d=3)$
The solution $a=d$ leads to planes intersecting in one point so it's been eliminated
Hope this helps
