Let \(L\), be the line \(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}\) and let \(S\) be the set of all points \((a, b, c)\…

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}=\lambda\)

\( x=2 \lambda-1, y=3 \lambda-1 \) & \(z=6 \lambda-3\)

\(\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}\)

\( x=2 \mu-1, y=3 \mu-1 \) & \(z=9\)

Let \(M\) is the point of intersection of both given lines

\(\Rightarrow 2\lambda −1=2\mu −1,3\lambda −1=3\mu −1,6\lambda −3=9\)

\(\Rightarrow \lambda =2=\mu\)

\(\Rightarrow \text{M }\left(3,5,9\right)\)

Now let point \(P\) be \(\left(2\text{ K}−1,3\text{ K}−1,6\text{ K}−3\right)\) on \(\text{L}\) such that \(\text{PM}=7\)

\(\Rightarrow \sqrt{{(2\text{K}−4)}^{2}+{(3\text{K}−6)}^{2}+{(6\text{K}−12)}^{2}}=7\)

\(\Rightarrow 49{\text{ K}}^{2}+196−196\text{K}=49\)

\(\Rightarrow {\text{K}}^{2}+4−4\text{K}=1\)

\(\Rightarrow {\text{K}}^{2}−4\text{K}+3=0\)

\(\Rightarrow \text{K}=1,3\)

So points are \(P\left(1,2,3\right)\) or \(P\left(5,8,15\right)\)

So sum of all co-ordinates of \(P = 34\)