JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given that

$$\vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$$a=λi^+j^​−k^

$$\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$$b=3i^−j^​+2k^

Also $(\vec{a}+\vec{b}+\vec{c})\times \vec{c}=\vec{0}$(a+b+c)×c=0

$$\Rightarrow k(\vec{a}+\vec{b})=\vec{c}$$⇒k(a+b)=c

$\vec{a}\cdot \vec{c}=-17$a⋅c=−17 and $\vec{b}\cdot \vec{c}=-20$b⋅c=−20

Now, Take $\vec{a}\cdot \vec{c}=-17$a⋅c=−17

$$\Rightarrow k(\lambda \hat{i}+\hat{j}-\hat{k})\cdot (\lambda \hat{i}+\hat{j}-\hat{k}+3\hat{i}-\hat{j}+2\hat{k})=-17$$⇒k(λi^+j^​−k^)⋅(λi^+j^​−k^+3i^−j^​+2k^)=−17

$$\Rightarrow k({\lambda }^{2}+3\lambda +0-1)=-17$$⇒k(λ2+3λ+0−1)=−17

$$\Rightarrow k({\lambda }^{2}+3\lambda -1)=-17....(1)$$⇒k(λ2+3λ−1)=−17....(1)

Similarly, on taking $\vec{b}\cdot \vec{c}=-20$b⋅c=−20, we get,

$$k(3\hat{i}-\hat{j}+2\hat{k})\cdot (\lambda \hat{i}+\hat{j}-\hat{k}+3\hat{i}-\hat{j}+2\hat{k})=-20$$k(3i^−j^​+2k^)⋅(λi^+j^​−k^+3i^−j^​+2k^)=−20

$$\Rightarrow k(3\lambda +9+2)=-20$$⇒k(3λ+9+2)=−20

$$\Rightarrow k(3\lambda +11)=-20...(2)$$⇒k(3λ+11)=−20...(2)

Now on solving equation (1)&(2) we get,

$$\Rightarrow 20{\lambda }^{2}+9\lambda -207=0$$⇒20λ2+9λ−207=0

$$\Rightarrow \lambda =3,\frac{-69}{20}$$⇒λ=3,20−69​

For $\lambda =3,k=-1$λ=3,k=−1

$$\Rightarrow \vec{c}=-1(\vec{a}+\vec{b})$$⇒c=−1(a+b)

$$\Rightarrow -((\lambda +3)\hat{i}+\hat{k})=-6\hat{i}-\hat{k}$$⇒−((λ+3)i^+k^)=−6i^−k^

$$\Rightarrow \vec{c}\times (\lambda \hat{i}+\hat{j}+\hat{k})=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1\end{matrix}|=\hat{i}-3\hat{j}+6\hat{k}$$⇒c×(λi^+j^​+k^)=∣i^−63​j^​01​k^−11​∣=i^−3j^​+6k^

$$\Rightarrow {|\vec{c}\times (\lambda \hat{i}+\hat{j}+\hat{k})|}^{2}=46$$⇒∣c×(λi^+j^​+k^)∣2=46

Hence this is the correct option.