Let $$Q$$ and $$R$$ be the feet of perpendiculars from the point $$P(a, a, a)$$ on the lines $$x = y, z = 1$$ and $$x = -y, z = -1$$ respectively. If $$\angle QPR$$ is a right angle, then $$12a^2$$ is equal to ________

Line $$x=y,z=1$$: point $$(t,t,1)$$. Line $$x=-y,z=-1$$: point $$(s,-s,-1)$$.

$$Q=(t,t,1)$$ is foot of perp from $$P(a,a,a)$$: direction of line is $$(1,1,0)$$. $$(a-t,a-t,a-1)\cdot(1,1,0)=0$$: $$2(a-t)=0$$, $$t=a$$. So $$Q=(a,a,1)$$.

$$R=(s,-s,-1)$$ is foot of perp from $$P(a,a,a)$$: direction $$(1,-1,0)$$. $$(a-s,a+s,a+1)\cdot(1,-1,0)=0$$: $$a-s-a-s=0$$, $$s=0$$. So $$R=(0,0,-1)$$.

$$\vec{PQ}=(0,0,1-a)$$, $$\vec{PR}=(-a,-a,-1-a)$$.

For right angle: $$\vec{PQ}\cdot\vec{PR}=0$$: $$(1-a)(-1-a)=0$$: $$-(1-a^2)=0$$: $$a^2=1$$.

$$12a^2=12$$.

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