The distance of the point \((1,3,-7)\) from the plane passing through the point \((1,-1,-1)\) having normal perpendicular to both the lines \(\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}\) and \(\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}\) is

Normal vector \(\hat{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -2 & 3 \\ 2 & -1 & -1\end{array}\right|\)
\(\begin{aligned}
& =\hat{\mathrm{i}}(2+3)-\hat{\mathrm{j}}(-1-6)+\hat{\mathrm{k}}(-1+4) \\
& =5 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}
\end{aligned}\)
Let \(\mathrm{A} \equiv(1,-1,-1)\)
\(\therefore \quad \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}\)
\(\therefore \quad\) Equation of the plane is
\(\begin{aligned}
& 5(x-1)+7(y+1)+3(z+1)=0 \\
& \Rightarrow 5 x+7 y+3 z+5=0
\end{aligned}\)
Distance of \((1,3,-7)\) from the above plane is \(d=\left|\frac{5(1)+7(3)+3(-7)+5}{\sqrt{25+49+9}}\right|=\frac{10}{\sqrt{83}}\) units