Given Information
We are given that $H = \operatorname { Span } \left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } \right\}$ and $K = \operatorname { Span } \left\{ \mathbf { v } _ { 3 } , \mathbf { v } _ { 4 } \right\} ,$ where \[ \mathbf { v } _ { 1 } = \left[ \begin{array} { l } { 5 } \\ { 3 } \\ { 8 } \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { l } { 1 } \\ { 3 } \\ { 4 } \end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } { 2 } \\ { - 1 } \\ { 5 } \end{array} \right] , \mathbf { v } _ { 4 } = \left[ \begin{array} { r } { 0 } \\ { - 12 } \\ { - 28 } \end{array} \right] \] Then $H$ and $K$ are subspaces of $\mathbb { R } ^ { 3 } .$ In fact, $H$ and $K$ are planes in $\mathbb { R } ^ { 3 }$ through the origin, and they intersect in a line through $\mathbf { 0 }$ . We have to find a nonzero vector $\mathbf { w }$ that generates that line.

Step-1:
since $H = \operatorname { span } \left\{ \mathbf { v } _ { 1 } , \mathbf { v } _ { 2 } \right\}$ and $K = \operatorname { span } \left\{ \mathbf { v } _ { 3 } , \mathbf { v } _ { 4 } \right\}$ intersect in a line, so $\mathrm { w }$ is in both $H$ and $K$. Therefore w can be be expressed as a linear combination of $v_1, v_2$:

$\mathbf { w } = c _ { 1 } \mathbf { v } _ { 1 } + c _ { 2 } \mathbf { v } _ { 2 }$ and $\mathbf { w } = c _ { 3 } \mathbf { v } _ { 3 } + c _ { 4 } \mathbf { v } _ { 4 }$ Thus, on substitution of vectors: \[\begin{array}{l} {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} = {c_3}{{\bf{v}}_3} + {c_4}{{\bf{v}}_4}\\ {c_1}{{\bf{v}}_1} + {c_2}{{\bf{v}}_2} - {c_3}{{\bf{v}}_3} - {c_4}{{\bf{v}}_4} = 0\\ {c_1}\left[ {\begin{array}{*{20}{l}} 5\\ 3\\ 8 \end{array}} \right] + {c_2}\left[ {\begin{array}{*{20}{l}} 1\\ 3\\ 4 \end{array}} \right] + {c_3}\left[ {\begin{array}{*{20}{c}} { - 2}\\ 1\\ { - 5} \end{array}} \right] + {c_4}\left[ {\begin{array}{*{20}{c}} 0\\ {12}\\ {28} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\]

Step-2:
Write the augmented matrix [A 0]: \[M = \left[ {\begin{array}{*{20}{c}} 5&1&{ - 2}&0&0\\ 3&3&1&{12}&0\\ 8&4&{ - 5}&{28}&0 \end{array}} \right]\] Row-Reduce the augmented Matrix: \[ \left[ \begin{array} { c c c c } { 1 } & { 0 } & { 0 } & { - \dfrac { 10 } { 3 } } & { 0 } \\ { 0 } & { 1 } & { 0 } & { \dfrac { 26 } { 3 } } & { 0 } \\ { 0 } & { 0 } & { 1 } & { - 4 } & { 0 } \end{array} \right] \]

Step-3:
System of equations: \[ c _ { 1 } - \dfrac { 10 } { 3 } c _ { 4 } = 0 , c _ { 2 } + \dfrac { 26 } { 3 } c _ { 4 } = 0 , c _ { 3 } - 4 c _ { 4 } = 0 \] Choosing $c_4 = t$ as a free variable: \[ c _ { 1 } = \dfrac { 10 t } { 3 } , c _ { 2 } = - \dfrac { 26 t } { 3 } , c _ { 3 } = 4 t \]

Step-4:
The general Solution: \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{c_1}}\\ {{c_2}}\\ {{c_3}}\\ {{c_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{{10}}{3}t}\\ { - \dfrac{{26}}{3}t}\\ {4t}\\ t \end{array}} \right]\\ = {c_4}\left[ {\begin{array}{*{20}{r}} {\dfrac{{10}}{3}}\\ { - \dfrac{{26}}{3}}\\ 4\\ 1 \end{array}} \right]\\ = \dfrac{{{c_4}}}{3}\left[ {\begin{array}{*{20}{c}} {10}\\ { - 26}\\ {12}\\ 3 \end{array}} \right] \end{array}\] So, the required vector w is: \[{\bf{w}} = 10\left[ {\begin{array}{*{20}{l}} 5\\ 3\\ 8 \end{array}} \right] - 26\left[ {\begin{array}{*{20}{l}} 1\\ 3\\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {24}\\ { - 48}\\ { - 24} \end{array}} \right]\] Therefore,

\[{\bf{w}} = \left[ {\begin{array}{*{20}{c}} {24}\\ { - 48}\\ { - 24} \end{array}} \right]\]