When the trial vector [vec{X}^{(1)}=left{begin{array}{l}1 1 1end{array} ight}] is used for the solution of the eigenvalue problem, [left[begin{array}{lll}1 & 1 & 2 1 & 2

When the trial vector

\[\vec{X}^{(1)}=\left\{\begin{array}{l}1 \\1 \\1\end{array}\right\}\]

is used for the solution of the eigenvalue problem,

\[\left[\begin{array}{lll}1 & 1 & 2 \\1 & 2 & 2 \\1 & 2 & 3\end{array}\right] \vec{X}=\lambda \vec{X}\]

the next trial vector, \(\vec{X}^{(2)}\), given by the matrix iteration method is

a. \(\left\{\begin{array}{l}3 \\ 5 \\ 6\end{array}\right\}\)

b. \(\left\{\begin{array}{l}1 \\ 1 \\ 1\end{array}\right\}\)

c. \(\left\{\begin{array}{l}3 \\ 3 \\ 3\end{array}\right\}\)