Line $L_1$ given by $y=x ; z=1$ can be expressed

$$ L_1: \frac{X}{1}=\frac{Y}{1}=\frac{z-1}{0} $$

$$ \begin{aligned} \frac{x}{1} & =\frac{y}{1}=\frac{z-1}{0}=\alpha (say)\\\\ \Rightarrow x & =\alpha, y=\alpha, z=1 \end{aligned} $$

Let the coordinates of $Q$ on $L_1$ be $(\alpha, \alpha, 1)$

Line $L_2$ given by $y=-x, z=-1$ can be expressed as

$L_2: \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=\beta $ (say)

$$ \Rightarrow \quad x=\beta, y=-\beta, z=-1 $$

Let the coordinates of $R$ on $L_2$ be $(\beta,-\beta,-1)$.

Direction ratios of $P Q$ are $\lambda-\alpha, \lambda-\alpha, \lambda-1$.

Now, $P Q \perp L_1$

$\begin{aligned} & \therefore 1(\lambda-\alpha)+1 \cdot(\lambda-\alpha)+0 \cdot(\lambda-1)=0 \\\\ & \Rightarrow \lambda=\alpha \\\\ & \therefore Q(\lambda, \lambda, 1)\end{aligned}$

Direction ratio of $P R$ are

$$ \lambda-\beta, \lambda+\beta, \lambda+1 \text {. } $$

Now, $P R \perp L_2$

$\begin{aligned} & \therefore 1(\lambda-\beta)+(-1)(\lambda+\beta)+0(\lambda+1)=0 \\\\ & \lambda-\beta-\lambda-\beta=0 \\\\ & \Rightarrow \beta=0 \\ & \end{aligned}$
$\therefore R(0,0,-1)$

Now, as $ \angle Q P R=90^{\circ}$

$$ \text { (as } a_1 a_2+b_1 b_2+c_1 c_2=0 \text {, } $$ if two lines with DR's $a_1, b_1, c_1 ; a_2, b_2, c_2$ are perpendicular)

$\begin{array}{cc}\therefore & (\lambda-\lambda)(\lambda-0)+(\lambda-\lambda)(\lambda-0) +(\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow (\lambda-1)(\lambda+1)=0 \\\\ &\Rightarrow \lambda=1 \text { or } \lambda=-1\end{array}$

$\lambda=1$, rejected as $P$ and $Q$ are different points.

$$ \Rightarrow \lambda=-1 $$