What is the equation of the plane passing through the point (1, 1,...

The direction ratios of the normal to the plane are proportional to the direction ratios of the line perpendicular to it. Thus, the normal vector has direction ratios $\langle 3, 2, 1 \rangle$.

The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal direction ratios $\langle a, b, c \rangle$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.

Substituting $(x_1, y_1, z_1) = (1, 1, 1)$ and $\langle a, b, c \rangle = \langle 3, 2, 1 \rangle$, we get:

$3(x - 1) + 2(y - 1) + 1(z - 1) = 0$

$3x - 3 + 2y - 2 + z - 1 = 0$

$3x + 2y + z = 6$

Answer: $3x + 2y + z = 6$