(1 point) Find a basis of the subspace of $R^{3}$ defined by the equation $-7 x_{1}+5 x_{2}-2 x_{3}=0$

Step 1 ：The given equation is a linear equation in three variables \(x_1\), \(x_2\), and \(x_3\). This equation defines a plane in \(R^3\). A basis for this plane (or subspace of \(R^3\)) can be found by finding two linearly independent solutions to the equation.

Step 2 ：We can find these solutions by setting two of the variables equal to 1 in turn and solving for the third variable. For example, if we set \(x_2 = 1\) and \(x_3 = 0\), we can solve for \(x_1\). Similarly, if we set \(x_1 = 0\) and \(x_3 = 1\), we can solve for \(x_2\). These two solutions will form a basis for the subspace.

Step 3 ：Solving the equation \(-7x_1 + 5 = 0\) gives us \(x_1 = \frac{5}{7}\).

Step 4 ：Solving the equation \(5x_2 - 2 = 0\) gives us \(x_2 = \frac{2}{5}\).

Step 5 ：Therefore, the two vectors that form a basis for the subspace defined by the equation \(-7 x_{1}+5 x_{2}-2 x_{3}=0\) are \(\left(\frac{5}{7}, 1, 0\right)\) and \(\left(0, \frac{2}{5}, 1\right)\).

Step 6 ：\(\boxed{\text{Final Answer: The basis of the subspace of } R^{3} \text{ defined by the equation } -7 x_{1}+5 x_{2}-2 x_{3}=0 \text{ is } \left\{\left(\frac{5}{7}, 1, 0\right), \left(0, \frac{2}{5}, 1\right)\right\}}\)