JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 4)
Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$$. Then $$\alpha+\beta+\gamma$$ is equal to :
16
20
18
14
Explanation
$$(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)$$
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DR of $$P Q$$ :
$$(2 \lambda-7,3 \lambda-6,5 \lambda-5)$$
DR of line : $$(2,3,5)$$
$$\begin{array}{ll} \therefore & 2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \\ \Rightarrow & \lambda=\frac{3}{2} \\ \therefore \quad & Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \\ \therefore \quad & (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text { is mid point of } P \& P^{\prime}\right) \\ & \alpha+\beta+y \equiv 14 \end{array}$$
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