JEE Main 2024 — Mathematics Vectors & 3D Geometry

Given:

$\frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}$0x−2​=2y​=−1z+3​ and $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}$1x−1​=−3y+3​=1z+1​

$$\Rightarrow \frac{x-2}{0}=\frac{y}{2}=\frac{z+3}{-1}=k(\text{let})$$⇒0x−2​=2y​=−1z+3​=k(let)

$$\Rightarrow x=2,y=2k,z=-k+3...(i)$$⇒x=2,y=2k,z=−k+3...(i)

Also, $\frac{x-1}{1}=\frac{y+3}{-3}=\frac{z+1}{1}=\lambda (\text{let})$1x−1​=−3y+3​=1z+1​=λ(let)

$$\Rightarrow x=\lambda +1,y=-3\lambda -3.z=\lambda -1...(ii)$$⇒x=λ+1,y=−3λ−3.z=λ−1...(ii)

Using $(i)\text{and}(ii)$(i)and(ii),

$$\Rightarrow 2=\lambda +1,2k=-3\lambda -3$$⇒2=λ+1,2k=−3λ−3

$$\Rightarrow \lambda =1,2k=-3-3$$⇒λ=1,2k=−3−3

$$\Rightarrow \lambda =1,k=-3$$⇒λ=1,k=−3

$$\Rightarrow x=2,y=-6,z=0$$⇒x=2,y=−6,z=0

So, the intersection point of two given lines is $(2,-6,0)$(2,−6,0).

Now, we know that centroid $G$G divides $MR$MR (median) in the ratio $1:2$1:2.

So, by using section formula, $G\equiv (\frac{1\times -1+2\times 2}{3},\frac{1\times 4+2\times 1}{3},\frac{1\times 2+2\times 2}{3})$G≡(31×−1+2×2​,31×4+2×1​,31×2+2×2​)

So, centroid is $G(1,2,2)$G(1,2,2).

Let $A$A be the point of intersection of given lines.

$$\Rightarrow AG=\sqrt{{(2-1)}^{2}+{(2+6)}^{2}+{(2-0)}^{2}}$$⇒AG=(2−1)2+(2+6)2+(2−0)2​

$$\Rightarrow AG=\sqrt{1+64+4}$$⇒AG=1+64+4​

$$\Rightarrow AG=\sqrt{69}$$⇒AG=69​