Consider the three planes
$$P_1: 3x + 15y + 21z = 9$$
$$P_2: x - 3y - z = 5$$, and
$$P_3: 2x + 10y + 14z = 5$$
Then, which one of the following is true?

We are given three planes: $$P_1: 3x + 15y + 21z = 9$$, $$P_2: x - 3y - z = 5$$, and $$P_3: 2x + 10y + 14z = 5$$.

Two planes are parallel if their normal vectors are proportional. The normal vectors are $$\vec{n_1} = (3, 15, 21)$$, $$\vec{n_2} = (1, -3, -1)$$, and $$\vec{n_3} = (2, 10, 14)$$.

Simplifying $$\vec{n_1}$$ by dividing by 3 gives $$(1, 5, 7)$$. Simplifying $$\vec{n_3}$$ by dividing by 2 gives $$(1, 5, 7)$$.

Since $$\vec{n_1}$$ and $$\vec{n_3}$$ are proportional (both reduce to $$(1, 5, 7)$$), the planes $$P_1$$ and $$P_3$$ are parallel. They are not identical since $$P_1$$ gives $$x + 5y + 7z = 3$$ while $$P_3$$ gives $$x + 5y + 7z = \frac{5}{2}$$.

The normal vector $$\vec{n_2} = (1, -3, -1)$$ is not proportional to $$(1, 5, 7)$$, so $$P_2$$ is not parallel to either $$P_1$$ or $$P_3$$.

The answer is Option 4: $$P_1$$ and $$P_3$$ are parallel.