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