System of equations based question in matrix form If $\begin{bmatrix}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\\4^2&5^2&6^2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\4\\10\\\lambda\end{bmatrix}$,then $\lambda=$
$(A)17\hspace{1cm}(B)18\hspace{1cm}(C)19\hspace{1cm}(D)20$

I tried solving this question.
$1^2x+2^2y+3^2z=1$
$2^2x+3^2y+4^2z=4$
$3^2x+4^2y+5^2z=10$
I solved this system to get $x=\frac{25}{8},y=\frac{-5}{2},z=\frac{7}{8}$
$4^2x+5^2y+6^2z=\lambda$
I substituted $x,y,z$ in the above equation to get $\lambda=19$.
But this method is long and tedious.My book gives one method but that i cannot understand.Please help me understand this.Book's method is not self explanatory.
Book's solution:
$(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$
$\lambda=1-3\times4+3\times 10=19$

OR if someone knows any better and elegant way to solve this question,please tell me,like using eigen values etc.Thanks.

I don't have a complete answer but I do notice that the equation $n^2-3(n+1)^2+3(n+2)^2=(n+3)^2$ shows that when $[x,y,z]=[1,-3,3]$ then the top three rows of the matrix times $[x,y,z]^T$ equals the (transpose of the) fourth row.  And it also equals $[1,4,10]^T$.  And the second equation is $[1,4,10]\times[1,-3,3]^T=\lambda$.  
