Verified Solution • 2025–26 Session moderate QJEEM24Ma4ea0: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4ea0 Solution: Let and be the feet of perpendiculars from the point on the… Let Q\mathrm{Q}Q and R\mathrm{R}R be the feet of perpendiculars from the point P(a,a,a)\mathrm{P}(a, a, a)P(a,a,a) on the lines x=y,z=1x=y, z=1x=y,z=1 and x=−y,z=−1x=-y, z=-1x=−y,z=−1 respectively. If ∠QPR\angle \mathrm{QPR}∠QPR is a right angle, then 12a212 a^212a2 is equal to _________. Step-by-Step Verified Solution amp;x1=y1=z−10=r→Q(r,r,1)amp;x1=y−1=z+10=k→R(k,−k,−1)amp;PQ‾=(a−r)i^+(a−r)j^+(a−1)k^amp;a=r+a−r=0amp;2a=2r→a=ramp;PR‾=(a−k)i+(a+k)j^+(a+1)k^amp;a−k−a−k=0⇒k=0amp;As,PQ⊥PRamp;(a−r)(a−k)+(a−r)(a+k)+(a−1)(a+1)=0amp;a=1 or −1amp;12a2=12\begin{aligned}& \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\& \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\& \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\& a=r+a-r=0 \\& 2 a=2 r \rightarrow a=r \\& \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\& a-k-a-k=0 \Rightarrow k=0 \\& A s, P Q \perp P R \\& (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\& a=1 \text { or }-1 \\& 12 a^2=12\end{aligned}amp;1x=1y=0z−1=r→Q(r,r,1)amp;1x=−1y=0z+1=k→R(k,−k,−1)amp;PQ=(a−r)i^+(a−r)j^+(a−1)k^amp;a=r+a−r=0amp;2a=2r→a=ramp;PR=(a−k)i+(a+k)j^+(a+1)k^amp;a−k−a−k=0⇒k=0amp;As,PQ⊥PRamp;(a−r)(a−k)+(a−r)(a+k)+(a−1)(a+1)=0amp;a=1 or −1amp;12a2=12 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring