JEE Main 2024 — 4 April, Shift 2 — Question 19

$\quad\overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})$

$$\begin{aligned} & \vec{a} \cdot \vec{d}=\lambda(\vec{b} \cdot \vec{a}+\vec{c} \cdot \vec{a}) & \end{aligned}$$

$1=\lambda(1+x+5)$

$1=\lambda(x+6)..(1)$

$|\overrightarrow{\mathrm{d}}|=1, \quad \frac{1}{\lambda}=\mathrm{x}+6$

$|\lambda(\vec{b}+\vec{c})|=1$

$|\lambda((x+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1$

$\lambda^{2}\left((x+2)^{2}+6^{2}+2^{2}\right)=1$

$\mathrm{x}^{2}+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^{2}$

$x^{2}+4 x+44=x^{2}+12 x+36$

$8 x=8, x=1$

$\begin{vmatrix}1&1&1\\2&4&-5\\x&2&3\end{vmatrix}=\left(\overrightarrow a\times\overrightarrow b\right).c$

$\begin{vmatrix}0&0&1\\-2&9&-4\\x-2&-1&3\end{vmatrix}=2-9\left(x-2\right)$

$=20-9 \mathrm{x}$ at $x=1$

$20-9=11$