JEE Main 2024 — Mathematics Vectors & 3D Geometry

Given,

Equation of line $x=y,z=1$x=y,z=1

$$\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r$$1x​=1y​=0z−1​=r

Now, let any point on the above line be $Q(r,r,1)$Q(r,r,1)

Similarly, for line $x=-y,z=-1$x=−y,z=−1 we get,

$$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k$$1x​=−1y​=0z+1​=k

So, any point on the above line will be $R(k,-k,-1)$R(k,−k,−1)

Now, the direction ratio of $\vec{PQ}=(a-r)\hat{i}+(a-r)\hat{j}+(a-1)\hat{k}$PQ​=(a−r)i^+(a−r)j^​+(a−1)k^

Now, finding foot of perpendicular from $P$P on line $\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}$1x​=1y​=0z−1​ using perpendicular condition we get,

$$a-r+a-r+0=0$$a−r+a−r+0=0

$$\Rightarrow 2a=2r$$⇒2a=2r

$$\Rightarrow a=r$$⇒a=r

Hence, point $Q(a,a,1)$Q(a,a,1)

And similarly, the direction ratio of

$$\vec{PR}=(a-k)i+(a+k)\hat{j}+(a+1)\hat{k}$$PR=(a−k)i+(a+k)j^​+(a+1)k^

And again, finding foot of perpendicular from $P$P on line $\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$1x​=−1y​=0z+1​ using perpendicular condition we get,

$$a-k-a-k=0\Rightarrow k=0$$a−k−a−k=0⇒k=0

Hence, point $R(0,0,-1)$R(0,0,−1)

Also, given $\vec{PQ}\perp \vec{PR}$PQ​⊥PR

Now, using perpendicular condition we get,

$$(a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0$$(a−r)(a−k)+(a−r)(a+k)+(a−1)(a+1)=0

$$\Rightarrow (a-a)(a-0)+(a-a)(a+0)+(a-1)(a+1)=0$$⇒(a−a)(a−0)+(a−a)(a+0)+(a−1)(a+1)=0

$$\Rightarrow (a-1)(a+1)=0$$⇒(a−1)(a+1)=0

$\Rightarrow a=1$⇒a=1 or $-1$−1

Hence, $12{a}^{2}=12$12a2=12