1. Write the two lines in parametric form

The line $x=y,\ z=1$ can be written as $L_1:(t,t,1).$ Its direction vector is $\vec d_1=(1,1,0).$

The line $x=-y,\ z=-1$ can be written as $L_2:(s,-s,-1).$ Its direction vector is $\vec d_2=(1,-1,0).$

Given point $P=(a,a,a).$

2. Find foot of perpendicular from $P$ to $L_1$

Let $Q=(t,t,1)\in L_1.$ Since $Q$ is the foot of the perpendicular, we must have $\overrightarrow{PQ}\cdot \vec d_1=0.$ Now $\overrightarrow{PQ}=Q-P=(t-a,t-a,1-a).$ So, $\overrightarrow{PQ}\cdot (1,1,0)=(t-a)+(t-a)=2(t-a)=0.$ Hence, $t=a.$ Therefore, $Q=(a,a,1).$

3. Find foot of perpendicular from $P$ to $L_2$

Let $R=(s,-s,-1)\in L_2.$ Since $R$ is the foot of the perpendicular, we must have $\overrightarrow{PR}\cdot \vec d_2=0.$ Now $\overrightarrow{PR}=R-P=(s-a,-s-a,-1-a).$ Thus, $\overrightarrow{PR}\cdot (1,-1,0)=(s-a)+(-1)(-s-a)=s-a+s+a=2s=0.$ So, $s=0.$ Therefore, $R=(0,0,-1).$

4. Use the condition $\angle QPR=90^\circ$

Since the angle at $P$ between $PQ$ and $PR$ is a right angle, $\overrightarrow{PQ}\cdot \overrightarrow{PR}=0.$

Now, $\overrightarrow{PQ}=Q-P=(0,0,1-a),$ $\overrightarrow{PR}=R-P=(-a,-a,-1-a).$ Hence, $\overrightarrow{PQ}\cdot \overrightarrow{PR}=0\cdot(-a)+0\cdot(-a)+(1-a)(-1-a)=0.$ So, $-(1-a)(1+a)=0$ $1-a^2=0$ $a^2=1.$ Therefore, $12a^2=12.$

5. Compare with stored answer

Derived answer: $12a^2=12$ Stored correct answer: $12$

They match.