Let $M$ and $N$ be foots of the perpendiculars drawn from the point $\mathrm{P}(a, a, a)$ on the lines $x-\mathrm{y}=0, \mathrm{z}=1$ and $x+\mathrm{y}=0, \mathrm{z}=-1$ respectively and if $\angle \mathrm{MPN}=90^{\circ}$ then $a^2=$

Given point: $\mathrm{P}(a, a, a)$.

For line $L_1$ with equations $x - y = 0$, $z = 1$, a general point is $\mathrm{M}(t, t, 1)$. The direction vector is $\vec{d_1} = (1, 1, 0)$.

The condition $\vec{PM} \perp \vec{d_1}$, where $\vec{PM} = (t - a, t - a, 1 - a)$, gives:

$2(t - a) = 0 \Rightarrow t = a$

Thus, $\mathrm{M} = (a, a, 1)$.

For line $L_2$ with $x + y = 0$, $z = -1$, a general point is $\mathrm{N}(s, -s, -1)$. The direction vector is $\vec{d_2} = (1, -1, 0)$.

The condition $\vec{PN} \perp \vec{d_2}$, where $\vec{PN} = (s - a, -s - a, -1 - a)$, gives:

$2s = 0 \Rightarrow s = 0$

Thus, $\mathrm{N} = (0, 0, -1)$.

Given $\angle \mathrm{MPN} = 90^\circ$, the vectors $\vec{PM} = (0, 0, 1 - a)$ and $\vec{PN} = (-a, -a, -1 - a)$ are perpendicular.

Their dot product is zero:

$(1 - a)(-1 - a) = 0 \Rightarrow -(1 - a^2) = 0$

This simplifies to $a^2 = 1$.

$\boxed{1}$