Let a line L passing through the point P (1, 1, 1) be perpendicular to the lines $$\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$$ and $$\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point S (1, 0, - 1) intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ___.

The line L passes through the point P(1, 1, 1) and is perpendicular to the given lines. The direction vectors of the given lines are required to find the direction vector of L.

The first line is given by $$\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$$, so its direction vector is $$\vec{d_1} = \langle 4, 1, 1 \rangle$$.

The second line is given by $$\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$$. Since the denominator for z is 0, the direction vector is $$\vec{d_2} = \langle 1, 1, 0 \rangle$$.

The direction vector of L, denoted $$\vec{d_L}$$, is perpendicular to both $$\vec{d_1}$$ and $$\vec{d_2}$$. Therefore, $$\vec{d_L} = \vec{d_1} \times \vec{d_2}$$.

Compute the cross product:

$$\vec{d_L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i} \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ 1 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 1 \\ 1 & 1 \end{vmatrix}$$

Calculate each determinant:

$$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = (1)(0) - (1)(1) = -1$$

$$\begin{vmatrix} 4 & 1 \\ 1 & 0 \end{vmatrix} = (4)(0) - (1)(1) = -1$$, so the j-component is $$-(-1) = 1$$

$$\begin{vmatrix} 4 & 1 \\ 1 & 1 \end{vmatrix} = (4)(1) - (1)(1) = 3$$

Thus, $$\vec{d_L} = \langle -1, 1, 3 \rangle$$.

The parametric equations for line L passing through P(1, 1, 1) with direction $$\langle -1, 1, 3 \rangle$$ are:

$$x = 1 - t$$

$$y = 1 + t$$

$$z = 1 + 3t$$

where t is a parameter.

L intersects the yz-plane where x = 0:

$$1 - t = 0 \implies t = 1$$

Substitute t = 1:

$$y = 1 + 1 = 2$$

$$z = 1 + 3(1) = 4$$

Thus, point Q is (0, 2, 4).

Another line parallel to L has the same direction vector $$\langle -1, 1, 3 \rangle$$ and passes through S(1, 0, -1). Its parametric equations are:

$$x = 1 - s$$

$$y = s$$

$$z = -1 + 3s$$

where s is a parameter.

This line intersects the yz-plane where x = 0:

$$1 - s = 0 \implies s = 1$$

Substitute s = 1:

$$y = 1$$

$$z = -1 + 3(1) = 2$$

Thus, point R is (0, 1, 2).

The points are P(1, 1, 1), Q(0, 2, 4), R(0, 1, 2), and S(1, 0, -1). For parallelogram PQRS, the adjacent sides from P are PQ and PS.

Vector $$\overrightarrow{PQ} = Q - P = \langle 0-1, 2-1, 4-1 \rangle = \langle -1, 1, 3 \rangle$$

Vector $$\overrightarrow{PS} = S - P = \langle 1-1, 0-1, -1-1 \rangle = \langle 0, -1, -2 \rangle$$

The area of parallelogram PQRS is the magnitude of the cross product of $$\overrightarrow{PQ}$$ and $$\overrightarrow{PS}$$.

Compute $$\overrightarrow{PQ} \times \overrightarrow{PS}$$:

$$\overrightarrow{PQ} \times \overrightarrow{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 3 \\ 0 & -1 & -2 \end{vmatrix} = \hat{i} \begin{vmatrix} 1 & 3 \\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix}$$

Calculate each determinant:

$$\begin{vmatrix} 1 & 3 \\ -1 & -2 \end{vmatrix} = (1)(-2) - (3)(-1) = -2 + 3 = 1$$

$$\begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} = (-1)(-2) - (3)(0) = 2$$, so the j-component is $$-2$$

$$\begin{vmatrix} -1 & 1 \\ 0 & -1 \end{vmatrix} = (-1)(-1) - (1)(0) = 1$$

Thus, $$\overrightarrow{PQ} \times \overrightarrow{PS} = \langle 1, -2, 1 \rangle$$

The magnitude is $$\sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$

Therefore, the area of parallelogram PQRS is $$\sqrt{6}$$, and the square of the area is $$(\sqrt{6})^2 = 6$$.