JEE Main 2024 — Vector Algebra Question with Solution

$\begin{array}{rl}& \overrightarrow{\mathrm{d}}=\lambda (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\\ & \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{d}}=\lambda (\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}\cdot \overrightarrow{\mathrm{a}})\\ & 1=\lambda (1+\mathrm{x}+5)\end{array}$ $1=\lambda (x+6)$ ...(1) $\begin{aligned} & |\overrightarrow{\mathrm{d}}|=1 \\ & |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \\ & |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \\ & \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 \\ & \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \\ & \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \\ & 8 \mathrm{x}=8, \mathrm{x}=1 \\ & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & -5 \\ \mathrm{x} & 2 & 3 \end{array}\right|=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \\ & \left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 & 9 & -4 \\ \mathrm{x}-2 & -1 & 3 \end{array}\right|=2-9(\mathrm{x}-2) \\ & =20-9 \mathrm{x} \\ & \text { at } \mathrm{x}=1 \\ & 20-9=11 \end{aligned}$