The perpendicular distance from the origin to the plane containing the two lines…
The perpendicular distance from the origin to the plane containing the two lines $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$, is
$\frac{11}{\sqrt{6}}$ units
$11 \sqrt{6}$ units
11 units
$6 \sqrt{11}$ units
Solution
Equation of the plane is
$\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-4 & z+4 \\
3 & 5 & 7 \\
1 & 4 & 7
\end{array}\right| \\
& \Rightarrow(x-1)(7)-(y-4)(14)+(z+4)(7)=0 \\
& \Rightarrow 7 x-14 y+7 z+77=0 \\
& \Rightarrow x-2 y+z+11=0
\end{aligned}$
$\therefore \quad$ Perpendicular distance from origin to the plane is $\left|\frac{0-2(0)+0+11}{\sqrt{1+4+1}}\right|=\frac{11}{\sqrt{6}}$ units