JEE Main 2022 — Mathematics Vectors & 3D Geometry

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

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

Let $\vec{c}=x\hat{i}+y\hat{j}+z\hat{k}$c=xi^+yj^​+zk^

So, $\vec{a}\times (\vec{b}+\vec{c})=\vec{0}\Rightarrow \vec{a}\times \vec{b}+\vec{a}\times \vec{c}=\vec{0}$a×(b+c)=0⇒a×b+a×c=0

$$\Rightarrow \vec{a}\times \vec{b}=-\vec{a}\times \vec{c}\Rightarrow \vec{a}\times \vec{b}=\vec{c}\times \vec{a}$$⇒a×b=−a×c⇒a×b=c×a

$$\Rightarrow |\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & 1 & 1\end{matrix}|=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & -2 & 3\end{matrix}|$$⇒∣i^11​j^​−21​k^31​∣=∣i^x1​j^​y−2​k^z3​∣

$$\Rightarrow -5\hat{i}+2\hat{j}+3\hat{k}=\hat{i}(3y+2z)+\hat{j}(z-3x)+\hat{k}(-2x-y)$$⇒−5i^+2j^​+3k^=i^(3y+2z)+j^​(z−3x)+k^(−2x−y)

On comparing we get,

$3y+2z=-5$3y+2z=−5 .....(i)

$z-3x=2$z−3x=2 ............(ii)

$-2x-y=3$−2x−y=3 ........(iii)

On solving equation (i), (ii) and (iii) we get,

$$x=\frac{-1}{6},y=\frac{-8}{3},z=\frac{3}{2}$$x=6−1​,y=3−8​,z=23​

So, $\vec{c}=\frac{-1}{6}\hat{i}+\frac{-8}{3}\hat{j}+\frac{3}{2}\hat{k}$c=6−1​i^+3−8​j^​+23​k^

So, $\vec{c}.\vec{a}=(\frac{-1}{6}\hat{i}+\frac{-8}{3}\hat{j}+\frac{3}{2}\hat{k}).(\hat{i}-2\hat{j}+3\hat{k})=\frac{29}{3}$c.a=(6−1​i^+3−8​j^​+23​k^).(i^−2j^​+3k^)=329​

So, $3\vec{c}.\vec{a}=29$3c.a=29

Note: This question appeared in JEE Main 2022, 29th june shift 2. The question was incorrect, so it is modified to make it correct.