If the foot is perpendicular from (1, 2, 3) to the line \(\frac{x+1}{2} = \frac{y-2}{5} = \frac{z-1}{1}\) is \(( a, \beta, \gamma)\), then find \(a + \beta + \gamma\)

Question

If the distances of the point \( (1,2,a) \) from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then \( a+b+c \) is equal to:

Answer 1

<div>The answer is option (B), which is 5.8.</div>

If the foot is perpendicular from (1, 2, 3) to the line \(\frac{x+1}{2} = \frac{y-2}{5} = \frac{z-1}{1}\) is \(( a, \beta, \gamma)\), then find \(a + \beta + \gamma\)

The Correct Option is B

Solution and Explanation

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