Final answer:
The equation of the line passing through the point (1, 0, 6) and perpendicular to the plane x - 3y + z = 5 is (x - 1) = -3(y - 0) = (z - 6).
Step-by-step explanation:
To find the equation of the line passing through the point (1, 0, 6) and perpendicular to the plane x - 3y + z = 5, we can use the normal vector of the plane, which is the coefficients of the x, y, and z terms. In this case, the normal vector is (1, -3, 1). To find the equation of the line, we can use the point-normal form of an equation of a line, which is given by:
(x - x0) / a = (y - y0) / b = (z - z0) / c
where (x0, y0, z0) is a point on the line and (a, b, c) is the normal vector. Plugging in the values (1, 0, 6) for (x0, y0, z0) and (1, -3, 1) for (a, b, c), we get:
(x - 1) / 1 = (y - 0) / -3 = (z - 6) / 1
So the equation of the line is (x - 1) = -3(y - 0) = (z - 6).