Let the equation of the plane passing through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$ be $$ax + by + cz = 65$$. Then the distance of the point $$(a, b, c)$$ from the plane $$2x + 2y - z + 16 = 0$$ is _____.

A plane passes through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and is parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$.

First, the direction of the second line (to which our plane is parallel) is found by computing the cross product of its normals. With $$\vec{n_1} = (1, 1, 2)$$ and $$\vec{n_2} = (2, 3, 1)$$, we have

$$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} = (1-6)\hat{i} - (1-4)\hat{j} + (3-2)\hat{k} = (-5, 3, 1)$$

Next, since the required plane passes through the first line, its equation can be written in the form

$$\lambda(x - 2y - z - 5) + \mu(x + y + 3z - 5) = 0$$

which expands to

$$(\lambda + \mu)x + (-2\lambda + \mu)y + (-\lambda + 3\mu)z - 5(\lambda + \mu) = 0$$

The normal vector of this plane is $$\vec{n} = (\lambda + \mu, -2\lambda + \mu, -\lambda + 3\mu)$$. Being parallel to the direction $$(-5, 3, 1)$$ requires that $$\vec{n} \cdot \vec{d} = 0$$, which gives

$$-5(\lambda + \mu) + 3(-2\lambda + \mu) + (-\lambda + 3\mu) = 0$$

$$-5\lambda - 5\mu - 6\lambda + 3\mu - \lambda + 3\mu = 0 \implies -12\lambda + \mu = 0 \implies \mu = 12\lambda$$

Choosing $$\lambda = 1$$ and $$\mu = 12$$ leads to the plane equation

$$13x + 10y + 35z - 65 = 0$$

In the standard form $$ax + by + cz = 65$$, we identify $$a = 13, b = 10, c = 35$$.

The distance of $$(a, b, c) = (13, 10, 35)$$ from the plane $$2x + 2y - z + 16 = 0$$ is computed by

$$d = \frac{|2(13) + 2(10) - 35 + 16|}{\sqrt{4 + 4 + 1}} = \frac{|26 + 20 - 35 + 16|}{3} = \frac{27}{3} = 9$$

The correct answer is 9.