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Mathematics · 2024

JEE Main · 30 January 2024, Shift 1 · Q16

Let (α, β, γ) be the foot of perpendicular from the point (1,2,3) on the line (x+3)/5=(y-1)/2=(z+4)/3. Then 19(α+β+γ) is equal to:

Let $\displaystyle (\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $\displaystyle (1,2,3)$ on the line $\displaystyle \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Then $\displaystyle 19(\alpha+\beta+\gamma)$ is equal to :
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