Let in a $\triangle A B C$, the length of the side $A C$ be 6 , the vertex $B$ be $(1,2,3)$ and the vertices…
- $17$
- $21$
- $56$
- $42$
Solution

$\begin{aligned} & \text { Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\ & \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4) \\ & \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \mid \overrightarrow{\mathrm{BM}}=7 \\ & \text { Area }=\frac{1}{2} \times 6 \times 7=21\end{aligned}$ *
Asked in: JEE Main 2025 (24 Jan Shift 1)