JEE Main 2024 — Vector Algebra Question with Solution

$\begin{array}{rl}& \vec{a}\times \vec{b}+\vec{a}\times \vec{c}+\vec{b}\times \vec{c}=(1,8,13)\\ & \vec{a}\times (\vec{a}\times \vec{b})+\vec{a}\times (\vec{a}\times \vec{c})+\vec{a}\times (\vec{b}\times \vec{c})\\ & =\vec{a}\times (\hat{i}+8\hat{j}+13\hat{k})\end{array}$
$\begin{array}{rl}& (\vec{a}\cdot \vec{b})\vec{a}-a^2\vec{b}+(\vec{a}\cdot \vec{c})\vec{a}-a^2\vec{c}+(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}=\vec{a}\times (\hat{i}+8\hat{j}+13\hat{k})\\ & \Rightarrow -26\vec{a}-29\vec{b}+13\vec{a}-29\vec{c}+13\vec{b}+26\vec{c}=\vec{a}\times (\hat{i}+8\hat{j}+13\hat{k})\end{array}$ $\begin{aligned} & \Rightarrow \quad-13 \overrightarrow{\mathrm{a}}-16 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \\ & \Rightarrow \quad-13 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-16 \mathrm{~b}^2-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\{\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})\} \cdot \overrightarrow{\mathrm{b}} \\ & \Rightarrow \quad(-13)(-26)-16(50)-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} 2 & -3 & 4 \\ 1 & 8 & 13 \\ 3 & 4 & -5 \end{array}\right| \\ & \Rightarrow \quad-462-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-396 \\ & \Rightarrow \quad \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-22 \end{aligned}$<br/>Hence$24-\vec{b} \cdot \vec{c}=46$