Answer
See below
Work Step by Step
$v_1, v_2$ and $v_3$ are linearly independent eigenvectors of $A$ corresponding to the eigenvalue $\lambda$,
and $c_1, c_2$ and $c_3$ are scalars.
Then,
$A(c_1v_1+c_2v_2+c_3v_2)\\
=c_1(Av_1)+c_2(Av_2)+c_3(Av_3)\\
=c_1(\lambda v_1)+c_2(\lambda v_2)+c_3(\lambda v_3)\\
=\lambda (c_1v_1+c_2v_2+c_3v_3)$
Hence, $c_1v_1+c_2v_2+c_3v_3$ is also an eigenvector of $A$ corresponding to the eigenvalue $\lambda$
