1. Key Concepts and Formulas

• Parameterization of a Line in 3D: A line in 3D space can be represented parametrically. For a line defined by two equations like $x=y, z=k$, a general point on the line can be written as $(t, t, k)$. Similarly, for $x=-y, z=k'$, a general point is $(s, -s, k')$. The direction vector of the line is obtained by taking the coefficients of the parameter (e.g., for $(t,t,k)$, the direction vector is $(1,1,0)$).
• Foot of Perpendicular from a Point to a Line: To find the foot of the perpendicular $Q$ from a point $P(x_0, y_0, z_0)$ to a line $L$ with direction vector $\vec{d}=(l,m,n)$, we first express a general point on the line $L$ parametrically, say $Q(\lambda)$. The vector $\vec{PQ}$ connecting $P$ to this general point must be perpendicular to the direction vector of the line $L$.
• Condition for Perpendicular Vectors: Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal) if and only if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.

2. Step-by-Step Solution

Step 1: Parameterize the lines and find their direction vectors. The given point is $P(a, a, a)$.

• Line 1: $x=y, z=1$. A general point on this line can be represented as $Q_g(t, t, 1)$. The direction vector of this line, $\vec{d_1}$, can be found by observing the change in coordinates with respect to $t$. Here, $\vec{d_1} = (1, 1, 0)$.

• Line 2: $x=-y, z=-1$. A general point on this line can be represented as $R_g(s, -s, -1)$. The direction vector of this line, $\vec{d_2}$, is $\vec{d_2} = (1, -1, 0)$.

Step 2: Find the coordinates of Q, the foot of the perpendicular from P to Line 1. Let $Q(t, t, 1)$ be the foot of the perpendicular from $P(a, a, a)$ to Line 1. The vector $\vec{PQ}$ is given by: $\vec{PQ} = (t-a, t-a, 1-a)$ Since $\vec{PQ}$ is perpendicular to Line 1, its dot product with the direction vector $\vec{d_1}=(1,1,0)$ must be zero: $\vec{PQ} \cdot \vec{d_1} = (t-a)(1) + (t-a)(1) + (1-a)(0) = 0$ $2(t-a) = 0 \implies t=a$ Substituting $t=a$ back into the coordinates of $Q_g$, we get the coordinates of $Q$: $Q = (a, a, 1)$

Step 3: Find the coordinates of R, the foot of the perpendicular from P to Line 2. Let $R(s, -s, -1)$ be the foot of the perpendicular from $P(a, a, a)$ to Line 2. The vector $\vec{PR}$ is given by: $\vec{PR} = (s-a, -s-a, -1-a)$ Since $\vec{PR}$ is perpendicular to Line 2, its dot product with the direction vector $\vec{d_2}=(1,-1,0)$ must be zero: $\vec{PR} \cdot \vec{d_2} = (s-a)(1) + (-s-a)(-1) + (-1-a)(0) = 0$ $s-a + s+a = 0$ $2s = 0 \implies s=0$ Substituting $s=0$ back into the coordinates of $R_g$, we get the coordinates of $R$: $R = (0, 0, -1)$

Step 4: Use the condition that $\angle QPR$ is a right angle. If $\angle QPR$ is a right angle, then the vectors $\vec{PQ}$ and $\vec{PR}$ must be perpendicular. First, let's find the components of $\vec{PQ}$ and $\vec{PR}$ using the calculated coordinates of $P, Q, R$: $P = (a, a, a)$ $Q = (a, a, 1)$ $R = (0, 0, -1)$ $\vec{PQ} = Q - P = (a-a, a-a, 1-a) = (0, 0, 1-a)$ $\vec{PR} = R - P = (0-a, 0-a, -1-a) = (-a, -a, -1-a)$ Now, set their dot product to zero: $\vec{PQ} \cdot \vec{PR} = (0)(-a) + (0)(-a) + (1-a)(-1-a) = 0$ $(1-a)(-1-a) = 0$ This simplifies to: $-(1-a)(1+a) = 0$ $-(1-a^2) = 0$ $a^2 - 1 = 0$ $a^2 = 1$

Step 5: Calculate the final required value. The problem asks for the value of $12a^2$. $12a^2 = 12(1) = 12$

3. Common Mistakes & Tips

• Incorrect Parameterization: Ensure the parametric equations correctly represent the given lines. For lines like $x=y, z=k$, remember that $z$ is constant, and $x$ and $y$ are equal (or related as specified).
• Dot Product Errors: Double-check the calculation of the dot product for both finding the foot of the perpendicular and for the angle condition. A sign error can lead to an incorrect result.
• Vector Definition: Always define vectors consistently (e.g., $\vec{PQ} = Q-P$). Reversing the order (e.g., $\vec{QP}$) would reverse the sign of the vector components, but the dot product for perpendicularity would still be zero.

4. Summary

This problem involved finding the feet of perpendiculars from a given point to two lines in 3D space, and then applying the condition that the angle between the two resulting vectors (from the given point to the feet of perpendiculars) is $90^\circ$. We parameterized each line, used the dot product condition for perpendicularity to find the coordinates of the feet of perpendiculars, and finally used the dot product condition for the angle between the two vectors $\vec{PQ}$ and $\vec{PR}$ to solve for $a^2$. The value of $a^2$ was found to be 1, leading to $12a^2 = 12$.

5. Final Answer

The final answer is $\boxed{12}$.