JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 10)
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 $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle A B C$ is:
42
17
56
21
Explanation
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$$\begin{aligned} & \text { Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\ & \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4) \\ & \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \mid \overrightarrow{\mathrm{BM}}=7 \\ & \text { Area }=\frac{1}{2} \times 6 \times 7=21 \end{aligned}$$
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