Equation of plane $(\pi)$ containing the points $(0,-5,-1),(1,-2,5),(-3,5,0)$
is $\left|\begin{array}{ccc}x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{ccc}x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1\end{array}\right|=0$
$\Rightarrow \quad 3 x+y-z+4=0$
$\Rightarrow \quad \mathbf{r}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+4=0$
$\Rightarrow \quad \mathbf{r}(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=4$
$\Rightarrow \mathbf{r}\left(\frac{-3}{\sqrt{11}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}}$
$\therefore$ Unit normal vector to plane $\pi$ is
$\hat{\mathbf{n}}=\frac{-3}{\sqrt{11}} \hat{\mathbf{i}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}$
$\therefore$ length of the projection of the unit normal vector $\hat{\mathbf{n}}$ to plane $\pi$ on the line $L$ is
$\left|\frac{\left(\frac{-3}{\sqrt{11}} \hat{\mathbf{i}} \frac{-\hat{\mathbf{j}}}{\sqrt{11}} \frac{+\hat{\mathbf{k}}}{\sqrt{11}}\right) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+25+36}}\right|$
$=\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}}$