JEE Main 2024 — Mathematics Vectors & 3D Geometry

Plotting the diagram of the given data we get,

Now, let $a,b,c$a,b,c be the DR's of the line perpendicular to $(-3\hat{i}+2\hat{k})+\lambda (2\hat{i}+3\hat{j}+5\hat{k})\text{and}(\hat{i}-2\hat{j}+\hat{k})+\mu (-\hat{i}+3\hat{j}+2\hat{k})$(−3i^+2k^)+λ(2i^+3j^​+5k^)and(i^−2j^​+k^)+μ(−i^+3j^​+2k^).

$$\Rightarrow 2a+3b+5c=0,-a+3b+2c=0$$⇒2a+3b+5c=0,−a+3b+2c=0

Now, solving 2a+3b+5c=0&-a+3b+2c=0 we get,

$$\Rightarrow \frac{a}{-9}=\frac{b}{-9}=\frac{c}{9}$$⇒−9a​=−9b​=9c​

$$\Rightarrow \frac{a}{1}=\frac{b}{1}=\frac{c}{-1}$$⇒1a​=1b​=−1c​

So, the equation will be,

$$\Rightarrow \frac{x-5}{1}=\frac{y+4}{1}=\frac{z-3}{-1}...(i)$$⇒1x−5​=1y+4​=−1z−3​...(i)

Let $R$R be the foot of perpendicular from $Q(0,2,-2)\text{to}(i)$Q(0,2,−2)to(i).

$$\Rightarrow \frac{x-5}{1}=\frac{y+4}{1}=\frac{z-3}{-1}=k$$⇒1x−5​=1y+4​=−1z−3​=k

$$\Rightarrow x=k+5,y=k-4,z=-k+3$$⇒x=k+5,y=k−4,z=−k+3

$$\Rightarrow R\equiv (k+5,k-4,-k+3)$$⇒R≡(k+5,k−4,−k+3)

So, DR's of $QR$QR are $(k+5,k-6,5-k)$(k+5,k−6,5−k).

Now, using perpendicular condition we get,

$$\Rightarrow 1(k+5)+1(k-6)+(-1)(5-k)=0$$⇒1(k+5)+1(k−6)+(−1)(5−k)=0

$$\Rightarrow k=2$$⇒k=2

$$\Rightarrow R\equiv (7,-2,1)$$⇒R≡(7,−2,1)

$$\Rightarrow QR=\sqrt{49+16+9}$$⇒QR=49+16+9​

$$\Rightarrow QR=\sqrt{74}$$⇒QR=74​