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I complete the solution in English language. Advanced MathIndicate the dimension of the subspace:a) Generated by\left[\begin{array}{rrr}6 & -1 & -1 \\18 & -3 & -3 \\18 & -3 & -3\end{array}\right] \mathbf{x}=\mathbf{0}Locating your answer in the following list:Linear space has dimension zero.Linear space has dimension one.Linear space has dimension two.Linear space has dimension three.Linear space has dimension four.R: [a]a) Let A=\left(\begin{array}{ccc}6 & -1 & -1 \\ 18 & -3 & -3 \\ 18 & -3 & -3\end{array}\right)Let X=\left(\begin{array}{lll}m_{1} & x_{2} & x_{3}\end{array}\right)^{\top}Then, A Y=0\Rightarrow\left(\begin{array}{ccc}6 & -1 & -1 \\18 & -3 & -3 \\18 & -3 & -3\end{array}\right)\left(\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right)=\left(\begin{array}{l}0 \\0 \\0\end{array}\right)Equiradently we can write\begin{array}{l}6 x_{1}-x_{2}-x_{3}=0 \\18 x_{1}-3 x_{2}-3 x_{3}=0 \\18 x_{1}-3 x_{2}-3 x_{3}=0\end{array}All equation in the abore ane some So, there are only ore equation,6 u_{1}-x_{2}-x_{3}=0Let m_{1}=e_{1}\begin{array}{l}\& x_{2}=c_{2} \\\therefore x_{2}=64+c_{2} \\\therefore x=\left(\begin{array}{l}x_{1} \\x_{2} \\x_{3}\end{array}\right)=\left(\begin{array}{c}c_{1} \\c_{2} \\64+c_{2}\end{array}\right) \\=C_{1}\left(\begin{array}{l}1 \\0 \\6\end{array}\right)+C_{2}\left(\begin{array}{l}0 \\1 \\1\end{array}\right) \\\end{array}So, the correct answer is(3) Livear space has dimension two i.e, (3) El espacio lineal tiene dimensión dos. ...