Question 1 — JEE Main 21 Jan 2026 (S I)

\[ \begin{aligned} & \text { let } \overrightarrow{\mathrm{c}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=x+y+z=4 \\ & \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 2 & 2 \\ x & y & z \end{array}\right| \\ & \Rightarrow 8 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=2(z-y) \hat{\mathrm{i}}-(-z-2 x) \hat{\mathrm{j}}+ \\ & z-y=4 \Rightarrow y=z-4 \\ & 2 x+z=7 \Rightarrow x=\frac{7-z}{2} ; 2 x+y=3 \\ & \text { From }(1) x+y+z=4 \Rightarrow \frac{7-z}{2}+z-4+ \\ & \Rightarrow 7-z+4 z-8=8 \Rightarrow 3 z=9 \Rightarrow z=3 \\ & x=\frac{7-z}{2}=2 ; y=z-4=-1 \\ & \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}=(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}|^{2}=1+1+25=27 \text { Answer }(4) \end{aligned} \]