JEE Main 2024 — Mathematics Vectors & 3D Geometry

$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}+(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \\ & \Rightarrow=\lambda(44 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =\lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\end{aligned}$​(a+b)×c−c×(−2a+3b)=0(a+b)×c+(−2a+3b)×c=0⇒(a+b)−2a+3b)×c=0⇒c=λ(4b−a)⇒=λ(44i^−4j^​+4k^−4i^+j^​−k^)=λ(40i^−3j^​+3k^)​ Now $\begin{aligned}
& (8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+33 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=1670 \
& \Rightarrow(41 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times \lambda=1670) \
& \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 \
& \text { so } \overrightarrow{\mathrm{c}}=40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \
& \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618
\end{aligned}$