$\text{If the equation of a line}\mathrm{ P}Q\text{ is }\frac{x+1}{2}=\frac{2-y}{5}=\frac{z+6}{7},\text{then the direction cosines of a line parallel to}\mathrm{ P}Q\text{ are}$

The correct option is:

$\frac{2}{\sqrt{78}},\frac{-5}{\sqrt{78}},\frac{7}{\sqrt{78}}$

Step-by-Step Explanation:

To find the direction cosines of a line parallel to the given line $PQ$, we first need to write the equation of $PQ$ in standard symmetrical form.
The standard symmetrical form of a straight line in three dimensions is:
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
where $a$, $b$, and $c$ are the direction ratios of the line.

The given equation of the line is:
$\frac{x+1}{2}=\frac{2-y}{5}=\frac{z+6}{7}$

Let's rearrange the second term, $\frac{2-y}{5}$, to match the standard form where the coefficient of $y$ is positive 1:
$\frac{2-y}{5}=\frac{-(y-2)}{5}=\frac{y-2}{-5}$

Substituting this back, the standard equation of the line $PQ$ is:
$\frac{x+1}{2}=\frac{y-2}{-5}=\frac{z+6}{7}$

From this equation, we can read off the direction ratios of the line $PQ$:
$a=2,b=-5,c=7$
Since any line parallel to $PQ$ has the same direction ratios (or proportional ones), we can use these direction ratios $(2,-5,7)$ to find its direction cosines.

The direction cosines $(l,m,n)$ are calculated from the direction ratios $(a,b,c)$ using the formulas:
$l=\frac{a}{\sqrt{a^2+b^2+c^2}},m=\frac{b}{\sqrt{a^2+b^2+c^2}},n=\frac{c}{\sqrt{a^2+b^2+c^2}}$

First, let's compute the value of the denominator term $\sqrt{a^2+b^2+c^2}$:
$\sqrt{2^2+{(-5)}^2+7^2}=\sqrt{4+25+49}=\sqrt{78}$

Now, substituting the values of $a$, $b$, and $c$ to find the direction cosines:
$l=\frac{2}{\sqrt{78}}$
$m=\frac{-5}{\sqrt{78}}$
$n=\frac{7}{\sqrt{78}}$

Thus, the direction cosines of a line parallel to $PQ$ are:
$\frac{2}{\sqrt{78}},\frac{-5}{\sqrt{78}},\frac{7}{\sqrt{78}}$