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

It is given that, 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.

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

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

$\Rightarrow$ $98\lambda = 98 \Rightarrow \lambda = 1$

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

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

$= \sqrt {36 + 9 + 4} = 7$ units