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:

• A. 102
• B. 101
• C. 99
• D. 100

Answer 1

The Correct Option is B

Given the line \(\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}\), we can parametrize it as:
\[x = 5t - 3, \quad y = 2t + 1, \quad z = 3t - 4\]

Let \(P(\alpha, \beta, \gamma) = (5t - 3, 2t + 1, 3t - 4)\) be the foot of the perpendicular from \(A = (1, 2, 3)\) to the line. The vector \(\overrightarrow{AP}\) is:

\[\overrightarrow{AP} = (5t - 4, 2t - 1, 3t - 7)\]

Since \(\overrightarrow{AP}\) is perpendicular to the line, we set up the dot product with the direction ratios \((5, 2, 3)\):

\[(5t - 4) \times 5 + (2t - 1) \times 2 + (3t - 7) \times 3 = 0\]

Expanding and solving:

\[38t - 43 = 0 \Rightarrow t = \frac{43}{38}\]

Substitute \(t = \frac{43}{38}\) to find \(\alpha\), \(\beta\), and \(\gamma\):

\[\alpha = 5t - 3 = \frac{101}{38}, \quad \beta = 2t + 1 = \frac{62}{19}, \quad \gamma = 3t - 4 = \frac{-23}{38}\]

Then,

\[\alpha + \beta + \gamma = \frac{101}{38} + \frac{124}{38} - \frac{23}{38} = \frac{202}{38} = \frac{101}{19}\]

Finally,

\[19(\alpha + \beta + \gamma) = 101\]