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Step-by-Step Solution
Step 1: Represent the lines in parametric form
The first line is given by the equations
x = y and z = 1.
Any point on this line can be written as
Q(r, r, 1),
where r is a parameter.
The second line is given by
x = -y and z = -1.
Any point on this line can be written as
R(k, -k, -1),
where k is another parameter.
Step 2: Write the coordinates of P, Q, and R
The point P is given as
P(a, a, a).
Hence,
\overrightarrow{PQ} = (r - a,\, r - a,\, 1 - a)
and
\overrightarrow{PR} = (k - a,\, -k - a,\, -1 - a).
Step 3: Condition for Q and R to be feet of perpendiculars
For Q to be the foot of the perpendicular from P onto the line x = y, z = 1,
vector \overrightarrow{PQ} must be perpendicular to the direction vector of that line.
Similarly, R must lie on x = -y, z = -1 and \overrightarrow{PR} must be perpendicular to its direction vector.
Through the perpendicularity conditions (or derived from a more detailed solving approach), we find
r = a and k = 0.
Thus,
Q = (a, a, 1)
and
R = (0, 0, -1).
Step 4: Impose the right angle condition at P
The angle \angle QPR is a right angle if
\overrightarrow{PQ} \cdot \overrightarrow{PR} = 0.
With
Q(a, a, 1) and R(0, 0, -1),
we have:
\overrightarrow{PQ} = Q - P = (a - a,\, a - a,\, 1 - a) = (0, 0, 1 - a),
\overrightarrow{PR} = R - P = (0 - a,\, 0 - a,\, -1 - a) = (-a, -a, -1 - a).
Their dot product is:
\overrightarrow{PQ} \cdot \overrightarrow{PR}
= (0)(-a) + (0)(-a) + (1 - a)(-1 - a).
Set this equal to 0:
(1 - a)(-1 - a) = 0.
Step 5: Solve for a
Simplify the expression:
(1 - a)(-1 - a)
= -(1 - a)(1 + a)
= -(1 - a^2).
So,
-(1 - a^2) = 0
\quad\Longrightarrow\quad
1 - a^2 = 0
\quad\Longrightarrow\quad
a^2 = 1
\quad\Longrightarrow\quad
a = \pm 1.
Step 6: Find 12a^2
Since a^2 = 1, we have
12 a^2 = 12 \times 1 = 12.
Final Answer:
12
