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.

You can solve this by row reducing the augmented matrix into echelon form. The augmented matrix is given by:
$$ \left[
    \begin{array}{ccc|c}
      1^2&2^2&3^2&1\\
      2^2&3^2&4^2&4\\
      3^2&4^2&5^2&10\\
      4^2&5^2&6^2&\lambda\\
    \end{array}
\right] $$
Row reducing leads to:
$$ \left[
    \begin{array}{ccc|c}
      1^2&2^2&3^2&1\\
      0&-7&-20&0\\
      0&0&\frac{8}{7}&1\\
      0&0&0&\lambda-19\\
    \end{array}
\right] $$
This system is only consistent if $\lambda - 19 = 0$ , therefore $\lambda = 19$.
