1. Key Concepts and Formulas

• Parametric Form of a Line: A line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ can be represented as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = t$. Any point on this line can be expressed as $(x_1+at, y_1+bt, z_1+ct)$.
• Intersection of a Line and a Plane: To find the intersection point, substitute the parametric coordinates of a general point on the line into the equation of the plane and solve for the parameter $t$.
• Foot of Perpendicular from a Point to a Line: If $Q$ is the foot of the perpendicular from point $R$ to line $L$, then the vector $\vec{RQ}$ is perpendicular to the direction vector of line $L$. Their dot product must be zero: $\vec{RQ} \cdot \vec{d_L} = 0$.
• Area of a Right-Angled Triangle: If a triangle $PQR$ is right-angled at $Q$, its area $\alpha$ can be calculated as $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} |\vec{QP}| |\vec{RQ}|$. Alternatively, the area can be found using the cross product: $\alpha = \frac{1}{2} |\vec{QP} \times \vec{QR}|$.

2. Step-by-Step Solution

Step 1: Parameterize the Line $L$ The given line $L$ is: $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ To represent any point on this line, we introduce a parameter, say $t$: $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}=t$ From this, the coordinates $(x, y, z)$ of any point on line $L$ can be expressed in terms of $t$: $x = 2t+1$ $y = -t-1$ $z = t+3$ So, a general point on line $L$ is $(2t+1, -t-1, t+3)$. The direction vector of line $L$ is $\vec{d_L} = (2, -1, 1)$.

Step 2: Find the Point $P$ (Intersection of Line $L$ and Plane) The point $P$ is the intersection of line $L$ and the plane $2x+y+3z=16$. Since $P$ lies on line $L$, its coordinates can be written as $(2t+1, -t-1, t+3)$ for some specific value of $t$. Since $P$ also lies on the plane, these coordinates must satisfy the plane's equation. Substitute the parametric coordinates of $P$ into the plane equation: $2(2t+1) + (-t-1) + 3(t+3) = 16$ Expand and simplify: $4t+2 - t-1 + 3t+9 = 16$ Combine like terms: $(4t - t + 3t) + (2 - 1 + 9) = 16$ $6t + 10 = 16$ $6t = 6$ $t = 1$ Now, substitute $t=1$ back into the parametric coordinates to find the coordinates of $P$: $P = (2(1)+1, -(1)-1, (1)+3) = (3, -2, 4)$ So, the point $P$ is $(3, -2, 4)$.

Step 3: Find the Point $Q$ (Foot of Perpendicular from $R$ to Line $L$) The point $R$ is given as $(1, -1, -3)$. Let $Q$ be the foot of the perpendicular from $R$ to line $L$. Since $Q$ lies on line $L$, its coordinates can be represented using the parametric form from Step 1, let's use a parameter $k$ to distinguish it from the parameter for $P$: $Q = (2k+1, -k-1, k+3)$ Now, we form the vector $\vec{RQ}$: $\vec{RQ} = Q - R = ((2k+1)-1, (-k-1)-(-1), (k+3)-(-3))$ $\vec{RQ} = (2k, -k, k+6)$ The direction vector of line $L$ is $\vec{d_L} = (2, -1, 1)$. Since $Q$ is the foot of the perpendicular from $R$ to $L$, the vector $\vec{RQ}$ must be perpendicular to the direction vector $\vec{d_L}$. Therefore, their dot product is zero: $\vec{RQ} \cdot \vec{d_L} = 0$ $(2k)(2) + (-k)(-1) + (k+6)(1) = 0$ $4k + k + k + 6 = 0$ $6k + 6 = 0$ $6k = -6$ $k = -1$ Substitute $k=-1$ back into the parametric coordinates of $Q$: $Q = (2(-1)+1, -(-1)-1, (-1)+3) = (-2+1, 1-1, -1+3) = (-1, 0, 2)$ So, the point $Q$ is $(-1, 0, 2)$.

Step 4: Calculate the Area of Triangle $PQR$ ($\alpha$) We have the coordinates of the three vertices: $P(3, -2, 4)$ $Q(-1, 0, 2)$ $R(1, -1, -3)$ Since $Q$ is the foot of the perpendicular from $R$ to the line $L$, and $P$ lies on line $L$, the line segment $RQ$ is perpendicular to the line segment $PQ$. This means that triangle $PQR$ is a right-angled triangle with the right angle at $Q$. The area $\alpha$ of a right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. Here, the base can be $|\vec{QP}|$ and the height can be $|\vec{RQ}|$.

First, calculate the vector $\vec{QP}$: $\vec{QP} = P - Q = (3 - (-1), -2 - 0, 4 - 2) = (4, -2, 2)$ Now, calculate the magnitude of $\vec{QP}$: $|\vec{QP}| = \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24}$ Next, calculate the vector $\vec{RQ}$: $\vec{RQ} = Q - R = (-1 - 1, 0 - (-1), 2 - (-3)) = (-2, 1, 5)$ Now, calculate the magnitude of $\vec{RQ}$: $|\vec{RQ}| = \sqrt{(-2)^2 + 1^2 + 5^2} = \sqrt{4 + 1 + 25} = \sqrt{30}$ The area $\alpha$ of triangle $PQR$ is: $\alpha = \frac{1}{2} |\vec{QP}| |\vec{RQ}| = \frac{1}{2} \sqrt{24} \sqrt{30}$ $\alpha = \frac{1}{2} \sqrt{24 \times 30} = \frac{1}{2} \sqrt{720}$

Step 5: Calculate $\alpha^2$ We need to find $\alpha^2$: $\alpha^2 = \left(\frac{1}{2} \sqrt{720}\right)^2 = \frac{1}{4} \times 720$ $\alpha^2 = 180$

3. Common Mistakes & Tips

• Parameter Consistency: Use distinct parameters (e.g., $t$ for $P$ and $k$ for $Q$) when finding multiple points on the same line to avoid confusion.
• Vector Direction: Ensure correct vector subtraction (e.g., $P-Q$ for $\vec{QP}$) to avoid sign errors.
• Geometric Interpretation: Recognize that the foot of the perpendicular from a point to a line creates a right-angled triangle with any other point on the line. This simplifies area calculation.
• Calculation Accuracy: Double-check arithmetic, especially with squares, square roots, and dot/cross products.

4. Summary

We first parameterized the given line to represent any point on it. Then, we used the plane equation to find the coordinates of point $P$. Next, we applied the condition for the foot of the perpendicular to find point $Q$. Finally, recognizing that triangle $PQR$ is right-angled at $Q$, we calculated the lengths of the sides $QP$ and $RQ$ and used them to find the area of the triangle. The square of the area, $\alpha^2$, was then calculated. Our calculations yield $\alpha^2 = 180$.

5. Final Answer The final answer is $\boxed{1}$.