Let Q and R be the feet of perpendiculars from the point P(a, a, a)...

Given,

Equation of line $x = y, z = 1$

$\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z - 1}{0} = r$

Now, let any point on the above line be $Q\left(r, r, 1\right)</p><p>Similarly, for line$x = - y, z = - 1$we get,</p><p>\dfrac{x}{1} = \dfrac{y}{- 1} = \dfrac{z + 1}{0} =$k$</p><p>So, any point on the above line will be$R\left(k, - k, - 1\right)$</p><p>Now, the direction ratio of$\overset{\to}{PQ} = \left(a - r\right)\overset{^}{i} + \left(a - r\right)\overset{^}{j} + \left(a - 1\right)\overset{^}{k}$</p><p>Now, finding foot of perpendicular from$P$on line$\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z - 1}{0} using perpendicular condition we get,

$a - r + a - r + 0 = 0$

$⇒ 2a = 2r$

$⇒ a = r$

Hence, point $Q\left(a, a, 1\right)$

And similarly, the direction ratio of 

\overset{\to}{PR} = \left(a - k\right)i + \left(a + k\right)\overset{^}{j} + \left(a + 1\right)\overset{^}{k}

And again, finding foot of perpendicular from $P$ on line $\dfrac{x}{1} = \dfrac{y}{- 1} = \dfrac{z + 1}{0} using perpendicular condition we get,</p><p>$a - k - a - k = 0 ⇒ k = 0$</p><p>Hence, point$R\left(0, 0, - 1\right)$</p><p>Also, given$\overset{\to}{PQ} ⊥ \overset{\to}{PR}$</p><p>Now, using perpendicular condition we get,</p><p>\left(a - r\right)$\left(a - k\right)$+ \left(a - r\right)$\left(a + k\right)$+ \left(a - 1\right)$\left(a + 1\right)$= 0$

$⇒ \left(a - a\right)$\left(a - 0\right)$+ \left(a - a\right)$\left(a + 0\right)$+ \left(a - 1\right)$\left(a + 1\right)$= 0$

$⇒ \left(a - 1\right)$\left(a + 1\right)$= 0$

$⇒ a = 1 or - 1$

Hence, $12{a}^{2} = 12$