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

Answer: (B)

101

Answer 2

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,

$19(\alpha + \beta + \gamma) = 101$