In a triangle with vertices$A\left( {1, - 1,2} \right)$, $B\left( {5, - 6,2} \right)$ and$C\left( {1,3, - 1} \right)$, find the altitude $n = \left| {BD} \right|$.
A. 5
B. 10
C. $5\sqrt 2 $
D. $10\sqrt 2 $

Hint:In this given question the vertices of a triangle are already given and we need to draw an altitude from point B which will on line AC of the triangle. Also in this $\Delta ABC$ there are two perpendicular AC and BD

Complete step by step Solution:
Given
We are given a $\Delta ABC$ with the following vertices$A\left( {1, - 1,2} \right)$, $B\left( {5, - 6,2} \right)$ and$C\left( {1,3, - 1} \right)$
On this $\Delta ABC$draw an altitude on a point between AC which connects to point B and that point being point D.
Let us assume that point D has following coordinates as $\left( {l,m,n} \right)$
Now, the line ac is $\dfrac{{\left( {x - 1} \right)}}{0} = \dfrac{{\left( {y + 1} \right)}}{4} = \dfrac{{\left( {z - 2} \right)}}{{ - 3}} = t$
From the above equation we can conclude that the point D on line AC will be $\left( {1,4t - 1,2 - 3t} \right)$
Therefore, point D$\left( {l,m,n} \right)$is $\left( {1,4t - 1,2 - 3t} \right)$
According to this
$
  l = 1 \\
  m = 4t - 1 \\
  n = 2 - 3t \\
 $
BD has \[l - 5,{\text{ }}m + 6,{\text{ }}n - 2\]
AC has \[0,4, - 3\]
In this $\Delta ABC$ there are two perpendicular AC and BD, by that we can conclude that
$4(m + 6) - 3(n - 2) = 0$
$4\left( {4t + 5} \right) - 3 - \left( {3t} \right) = 0$
$t = \dfrac{{ - 4}}{5}$
From the above equation we can find out the coordinates of point D$\left( {l,m,n} \right)$
$l = 4t - 1$
\[ = 4\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{ - 21}}{5}\]
$m = 1$
 $n = 2 - 3t$
$ = 2 - 3\left( {\dfrac{{ - 4}}{5}} \right) = \dfrac{{22}}{5}$
Thus point D is $\left( {1,\dfrac{{ - 21}}{5},\dfrac{{22}}{5}} \right)$
By the above information we can say find altitude $n = \left| {BD} \right|$
$n = \left| {BD} \right| = \sqrt {{1^2} + {{\left( {\dfrac{{21}}{5}} \right)}^2} + {{\left( {\dfrac{{22}}{5}} \right)}^2}} $
$ = \sqrt {25} = 5$
Hence, the correct option is A.

Note: In this we have to draw a vertex in $\Delta ABC$ and then find out the coordinates of that point. Also, we have to assume the point coordinate and make it into an equation form which makes it easy to derive an answer. We also need to know how to find BD with the help of the coordinates given.