Find the distance of a point $(2, 4, -1)$ from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}$.

$7$ units

The correct answer is Option (1) → $7$ units ##

We have, equation of the line as $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda$

$\Rightarrow x = \lambda - 5, \ y = 4\lambda - 3, \ z = 6 - 9\lambda$

Let the coordinates of $L$ be $(\lambda - 5, 4\lambda - 3, 6 - 9\lambda)$, then Dr's of $PL$ are $(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$.

Also, the direction ratios of given line are proportional to $1, 4, -9$.

Since, $PL$ is perpendicular to the given line.

$∴(\lambda - 7) \cdot 1 + (4\lambda - 7) \cdot 4 + (7 - 9\lambda) \cdot (-9) = 0$

$⇒\lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0$

$⇒98\lambda - 98 = 0 ⇒\lambda = 1$

So, the coordinates of $L$ are $(-4, -1, -3)$.

$∴$ Required distance, $PL = \sqrt{(-4 - 2)^2 + (-1 - 4)^2 + (-3 + 1)^2} = \sqrt{(-6)^2 + (-5)^2 + (-2)^2}$

$= \sqrt{36 + 25 + 4} \quad \text{No — as per image: } \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \text{ units}$