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Step-by-Step Solution
Step 1: Express the line in parametric form
The given line is
$$\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{-1} = \lambda \quad (\text{say}).$$
From this, we can write the parametric equations:
$$x = 2\lambda - 1,\quad y = 5\lambda + 1,\quad z = -\lambda - 1.$$
Step 2: Assume foot of the perpendicular
Let the foot of the perpendicular from the point
$A(2,\,0,\,5)$ onto the line be
$P(\alpha,\beta,\gamma)$. In terms of $\lambda$, this point is
$$P(2\lambda - 1,\; 5\lambda + 1,\; -\lambda - 1).$$
So we identify
$$\alpha = 2\lambda - 1,\quad \beta = 5\lambda + 1,\quad \gamma = -\lambda - 1.$$
Step 3: Use perpendicularity condition
The direction vector of the given line is
$$\overrightarrow{b} = \langle 2,\; 5,\; -1\rangle.$$
If $P$ is the foot of the perpendicular, then
$$\overrightarrow{PA} \cdot \overrightarrow{b} = 0,$$
where
$$\overrightarrow{PA} = \bigl\langle 2 - (2\lambda - 1),\; 0 - (5\lambda + 1),\; 5 - (-\lambda - 1)\bigr\rangle.$$
Simplifying,
$$\overrightarrow{PA} = \langle 3 - 2\lambda,\; -5\lambda - 1,\; 6 + \lambda\rangle.$$
Step 4: Compute the dot product and set it to zero
The dot product
$\overrightarrow{PA} \cdot \overrightarrow{b} = 0$ gives:
\[
\langle 3 - 2\lambda,\; -5\lambda - 1,\; 6 + \lambda\rangle
\cdot \langle 2,\; 5,\; -1\rangle = 0.
\]
Hence,
\[
2(3 - 2\lambda) + 5(-5\lambda - 1) + (-1)(6 + \lambda) = 0.
\]
Step 5: Solve for $\lambda$
Expanding and simplifying:
\[
2(3 - 2\lambda) = 6 - 4\lambda,\quad
5(-5\lambda - 1) = -25\lambda - 5,\quad
-1(6 + \lambda) = -6 - \lambda.
\]
Summing:
\[
(6 - 4\lambda) + (-25\lambda - 5) + (-6 - \lambda) = 0
\;\;\Longrightarrow\;\; 6 - 4\lambda - 25\lambda - 5 - 6 - \lambda = 0.
\]
\[
-30\lambda - 5 = 0
\;\;\Longrightarrow\;\; \lambda = -\frac{1}{6}.
\]
Step 6: Find the coordinates of the foot of perpendicular
Substitute $\lambda = -\tfrac{1}{6}$ into the parametric form:
\[
\alpha = 2\left(-\frac{1}{6}\right) - 1 = -\frac{1}{3} - 1 = -\frac{4}{3},
\]
\[
\beta = 5\left(-\frac{1}{6}\right) + 1 = -\frac{5}{6} + 1 = \frac{1}{6},
\]
\[
\gamma = -\left(-\frac{1}{6}\right) - 1 = \frac{1}{6} - 1 = -\frac{5}{6}.
\]
Hence,
$$P(\alpha,\beta,\gamma) = \bigl(-\tfrac{4}{3},\,\tfrac{1}{6},\,-\tfrac{5}{6}\bigr).$$
Step 7: Evaluate each given ratio and identify the incorrect one
$\displaystyle \frac{\alpha}{\beta}
= \frac{-\tfrac{4}{3}}{\tfrac{1}{6}}
= -\frac{4}{3} \times \frac{6}{1}
= -8.
\quad\checkmark\;\; (\text{Correct})$
$\displaystyle \frac{\alpha \,\beta}{\gamma}
= \frac{\left(-\tfrac{4}{3}\right)\left(\tfrac{1}{6}\right)}{-\tfrac{5}{6}}
= \frac{-\tfrac{4}{18}}{-\tfrac{5}{6}}
= \frac{-\tfrac{2}{9}}{-\tfrac{5}{6}}
= \frac{-2}{9} \times \frac{-6}{5}
= \frac{12}{45}
= \frac{4}{15}.
\quad\checkmark\;\; (\text{Correct})$
$\displaystyle \frac{\beta}{\gamma}
= \frac{\tfrac{1}{6}}{-\tfrac{5}{6}}
= \frac{1}{6} \times \left(-\frac{6}{5}\right)
= -\frac{1}{5}.
\quad \text{Given as } -5 \text{ (which is incorrect).}$
$\displaystyle \frac{\gamma}{\alpha}
= \frac{-\tfrac{5}{6}}{-\tfrac{4}{3}}
= \left(-\frac{5}{6}\right)\Big/\left(-\frac{4}{3}\right)
= \left(-\frac{5}{6}\right)\times\left(-\frac{3}{4}\right)
= \frac{15}{24}
= \frac{5}{8}.
\quad\checkmark\;\; (\text{Correct})$
Conclusion:
The statement
$\displaystyle \frac{\beta}{\gamma} = -5$
is the one that is NOT correct. In reality,
$\displaystyle \frac{\beta}{\gamma} = -\frac{1}{5}.$
Thus, the incorrect statement is
$\displaystyle \frac{\beta}{\gamma}=-5.$
