Find an equation of the plane through the origin which is parallel to the plane x − 6y + 4z = 7.

Linear equations in three-dimensional space is characterized by the standard formula: ax + by + cz  + d = 0, where a, b and c are coefficients and d is a constant. Three-dimensional space is different than two-dimensional because you don't base it on the slope of the line. The basis of two lines being parallel in three-dimensional space is the determinant of the matrix being zero. The a coefficients are in the first column, b in the second, c in the third and the constant in the fourth column. From the given equation, a = 1, b = --6, c = 4 and d = 7. Then, let the equation parallel to this would be unknowns x, y, z and k.

   1   -6   4   7
   x    y    z   k

To find the determinant on the matrix, cross multiply the diagonals. Find the sum of the diagonal facing down first, then subtract to this the sum of all the cross-multiplied terms of diagonals up.

(y + -6z + 4k) - (-6x + 4y + 7z) = 0
y - 6z + 4k + 6x + -4y + 7z = 0
6x  -5y  -13z  +4k = 0

This is the general form of the parallel line. Since it passes the origin, let's substitute the coordinates (0,0,0) to the x,y and z variables, respectively. Then, we can determine the value of k.

6(0) - 5(0) - 13(0) + 4k = 0
k = 0

Therefore, the expression for the parallel line is: 6x - 5y - 13z = 0.