Let the line L be $\frac{x-1}{1}=\frac{y-4}{3}=\frac{z-7}{5}$ and foot of perpendicular from (1,-2,-1) to L is $(\alpha ,\beta ,\gamma )$, then $\alpha +\beta +\gamma$ is :

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YESPYQ · Accepted Answer

Correct answer: d) $\frac{-102}{35}$. $\frac{x-1}{1}=\frac{y-4}{3}-\frac{z-7}{5}=\lambda (l\mathrm{et})\ B(\lambda +1,3\lambda +4,5\lambda +7)\ DR's\text{of}AB=⟨\lambda ,3\lambda +6,5\lambda +8⟩\ Now,\lambda +9\lambda +18+25\lambda +40=0\ \lambda =\frac{-58}{35}\ B=(\frac{-58}{35}+1,3\left(\frac{-58}{35}\right)+4,5\left(\frac{-58}{35}\right)+7)\ =\left(\frac{-23}{35},\frac{-34}{35},\frac{-45}{35}\right)=\left(\alpha ,\beta ,\gamma \right)\ \mathrm{Now},\alpha +\beta +\gamma =\frac{-23-34-45}{35}=\frac{-102}{35}\ \$

Let the line L be \(\frac{x-1}{1}=\frac{y-4}{3}=\frac{z-7}{5}\) and foot of perpendicular from (1,-2,-1) to L is \((\alp…

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