Question

Let $(\alpha, \beta, \gamma)$ be the point $(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to

• A. 16
• B. 18
• C. 14
• D. 20

Answer 1

The Correct Option is C

Let \( P(8, 5, 7) \), \( \vec{a} = \langle 2, 3, 5 \rangle \), \( F(2\lambda + 1, 3\lambda - 1, 5\lambda + 2) \).

The vector \( \overrightarrow{PF} \) is:

\[ \overrightarrow{PF} = \langle 2\lambda - 7, 3\lambda - 6, 5\lambda - 5 \rangle. \]

Since \( \overrightarrow{PF} \cdot \vec{a} = 0 \), we have:

\[ (2\lambda - 7)(2) + (3\lambda - 6)(3) + (5\lambda - 5)(5) = 0. \]

Simplify:

\[ 4\lambda - 14 + 9\lambda - 18 + 25\lambda - 25 = 0, \] \[ 38\lambda - 57 = 0 \implies \lambda = \frac{3}{2}. \]

The coordinates of \( F \) are:

\[ F = (2\lambda + 1, 3\lambda - 1, 5\lambda + 2) = \left( 4, \frac{7}{2}, \frac{19}{2} \right). \]

The coordinates of \( Q(\alpha, \beta, \gamma) \) are:

\[ \frac{\alpha + 8}{2} = 2\lambda + 1, \quad \frac{\beta + 5}{2} = 3\lambda - 1, \quad \frac{\gamma + 7}{2} = 5\lambda + 2. \]

Solving each equation:

\[ \alpha = 4\lambda - 6, \quad \beta = 6\lambda - 7, \quad \gamma = 10\lambda - 3. \]

Substitute \( \lambda = \frac{3}{2} \):

\[ \alpha = 4\left(\frac{3}{2}\right) - 6 = 6 - 6 = 0, \] \[ \beta = 6\left(\frac{3}{2}\right) - 7 = 9 - 7 = 2, \] \[ \gamma = 10\left(\frac{3}{2}\right) - 3 = 15 - 3 = 12. \]

Finally, compute \( \alpha + \beta + \gamma \):

\[ \alpha + \beta + \gamma = 0 + 2 + 12 = 14. \]

Final Answer: 14