1. Key Concepts and Formulas

• Equation of a Line in Symmetric Form: A line passing through point $(x_0, y_0, z_0)$ with direction vector $(a, b, c)$ is given by $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$. Any general point on the line can be represented by introducing a parameter, typically $\lambda$.
• Vector Perpendicularity (Orthogonality): Two vectors are perpendicular if and only if their dot product is zero. If $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$, then $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 = 0$.
• Foot of the Perpendicular: The foot of the perpendicular from a point $P$ to a line $L$ is a point $Q$ on the line $L$ such that the vector $\vec{PQ}$ is perpendicular to the direction vector of the line $L$.

2. Step-by-Step Solution

Step 1: Represent a General Point on the Line The given line is in symmetric form: $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{-1}$ To find the coordinates of any point $Q$ on this line, we introduce a parameter $\lambda$: $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{-1} = \lambda$ From this, we can express the coordinates of $Q(x, y, z)$ in terms of $\lambda$:

• $x + 1 = 2\lambda \implies x = 2\lambda - 1$
• $y - 1 = 5\lambda \implies y = 5\lambda + 1$
• $z + 1 = -\lambda \implies z = -\lambda - 1$ So, any general point on the line is $Q(2\lambda - 1, 5\lambda + 1, -\lambda - 1)$. This point $Q$ represents the foot of the perpendicular, $(\alpha, \beta, \gamma)$.

Step 2: Identify the Given Point and the Direction Vector of the Line The given point from which the perpendicular is drawn is $P(2, 0, 5)$. The direction vector of the line, $\vec{d}$, can be directly read from the denominators of the symmetric equation: $\vec{d} = (2\hat{i} + 5\hat{j} - \hat{k})$

Step 3: Form the Vector $\vec{PQ}$ The vector $\vec{PQ}$ connects the point $P(2, 0, 5)$ to the general point $Q(2\lambda - 1, 5\lambda + 1, -\lambda - 1)$ on the line. $\vec{PQ} = Q - P = ((2\lambda - 1) - 2, (5\lambda + 1) - 0, (-\lambda - 1) - 5)$ $\vec{PQ} = (2\lambda - 3, 5\lambda + 1, -\lambda - 6)$

Step 4: Apply the Orthogonality Condition Since $Q$ is the foot of the perpendicular from $P$ to the line, the vector $\vec{PQ}$ must be perpendicular to the direction vector of the line, $\vec{d}$. Therefore, their dot product must be zero: $\vec{PQ} \cdot \vec{d} = 0$ Substitute the components of $\vec{PQ}$ and $\vec{d}$: $(2\lambda - 3)(2) + (5\lambda + 1)(5) + (-\lambda - 6)(-1) = 0$

Step 5: Solve for the Parameter $\lambda$ Now, we solve the equation obtained from the dot product: $(4\lambda - 6) + (25\lambda + 5) + (\lambda + 6) = 0$ Combine the terms with $\lambda$ and constant terms: $(4\lambda + 25\lambda + \lambda) + (-6 + 5 + 6) = 0$ $30\lambda + 5 = 0$ $30\lambda = -5$ $\lambda = -\frac{5}{30} = -\frac{1}{6}$

Step 6: Find the Coordinates of the Foot of the Perpendicular Substitute the value of $\lambda = -\frac{1}{6}$ back into the expressions for the coordinates of point $Q(\alpha, \beta, \gamma)$: $\alpha = 2\lambda - 1 = 2\left(-\frac{1}{6}\right) - 1 = -\frac{1}{3} - 1 = -\frac{4}{3}$ $\beta = 5\lambda + 1 = 5\left(-\frac{1}{6}\right) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$ $\gamma = -\lambda - 1 = -\left(-\frac{1}{6}\right) - 1 = \frac{1}{6} - 1 = -\frac{5}{6}$ So, the foot of the perpendicular is $(\alpha, \beta, \gamma) = \left(-\frac{4}{3}, \frac{1}{6}, -\frac{5}{6}\right)$.

Step 7: Check the Given Options Now we evaluate each option using $\alpha = -\frac{4}{3}$, $\beta = \frac{1}{6}$, and $\gamma = -\frac{5}{6}$ to find which statement is NOT correct.

(A) $\frac{\alpha}{\beta}=-8$ $\frac{\alpha}{\beta} = \frac{-4/3}{1/6} = -\frac{4}{3} \times \frac{6}{1} = -4 \times 2 = -8$ This statement is Correct.

(B) $\frac{\alpha \beta}{\gamma}=\frac{4}{15}$ $\frac{\alpha \beta}{\gamma} = \frac{\left(-\frac{4}{3}\right)\left(\frac{1}{6}\right)}{-\frac{5}{6}} = \frac{-\frac{4}{18}}{-\frac{5}{6}} = \frac{-\frac{2}{9}}{-\frac{5}{6}} = \left(-\frac{2}{9}\right) \times \left(-\frac{6}{5}\right) = \frac{12}{45} = \frac{4}{15}$ This statement is Correct.

(C) $\frac{\beta}{\gamma}=-5$ $\frac{\beta}{\gamma} = \frac{1/6}{-5/6} = -\frac{1}{5}$ The option states $\frac{\beta}{\gamma}=-5$. Since $-\frac{1}{5} \neq -5$, this statement is NOT Correct.

(D) $\frac{\gamma}{\alpha}=\frac{5}{8}$ $\frac{\gamma}{\alpha} = \frac{-5/6}{-4/3} = \left(-\frac{5}{6}\right) \times \left(-\frac{3}{4}\right) = \frac{15}{24} = \frac{5}{8}$ This statement is Correct.

Based on our calculations, the statement in option (C) is the one that is NOT correct.

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when calculating the dot product and substituting $\lambda$ values. A single sign error can lead to a completely different result.
• Fraction Arithmetic: Ensure accurate manipulation of fractions during substitution and simplification. Double-check multiplication and division of fractions.
• Parameterization: Make sure to correctly parameterize the line from its symmetric form to avoid errors in defining the general point $Q$.
• Orthogonality Condition: Remember that the vector from the point to the foot of the perpendicular is orthogonal to the direction vector of the line, not necessarily the line itself.

4. Summary

We found the foot of the perpendicular $(\alpha, \beta, \gamma)$ by first parameterizing a general point on the line. Then, we formed a vector connecting the given point $P$ to this general point $Q$. By applying the orthogonality condition (dot product of $\vec{PQ}$ and the line's direction vector equals zero), we solved for the parameter $\lambda$. Substituting $\lambda$ back into the parameterized coordinates gave us the specific coordinates of the foot of the perpendicular. Finally, we checked each given option with these coordinates to identify the statement that was NOT correct. Our calculations show that option (C) is the incorrect statement.

5. Final Answer

The final answer is $\boxed{A}$.