Find the foot of the perpendicular from the point P(1,2,3) on the line x-6/3=y-7/2=z-7/-2 ? Also obtain the equation of the plane containing the line and the point (1,2,3).

Kindly provide the solution! Also, please explain how to find the direction ratios when the foot of the perpendicular lies on a line?

The given equation of the plane is $x-\frac{6}{3}=y-\frac{7}{2}=z-\left(\frac{7}{-2}\right)$
This equation can be rewritten as;
$$\begin{gathered} x-2=y-\frac{7}{2}=z+\frac{7}{2} \\ \Rightarrow x-2=y-\frac{7}{2}-\frac{7}{2}-z=0 \\ \Rightarrow x-2=y-z-7 \\ \Rightarrow x-y+z+5=0 \end{gathered}$$
Let $Q\left(a,b,c\right)$ be the point of the foot of the perpendicular drawn from the point $P\left(1,2,3\right)$
Since $Q\left(a,b,c\right)$ lies on the plane $x-y+z+5=0$ we get;
$a-b+c+5=0$..........(i)
The direction ratios of the normal to the given plane is 1, -1, 1.
The direction ratios of PQ is $\left(a-1\right),\left(b-2\right),\left(c-3\right)$.

Since, PQ is normal to the given plane , we have
$\frac{a-1}{1}=\frac{b-2}{-1}=\frac{c-3}{1}=k$ (say)
$$\begin{gathered} a-1=k, b-2=-k, c-3=k \\ \Rightarrow a=\left(k+1\right); b=\left(2-k\right); c=\left(k+3\right) \end{gathered}$$
Thus from (i) we have;
$$\begin{gathered} k+1-\left(2-k\right)+\left(k+3\right)+5=0 \\ \Rightarrow k+1-2+k+k+3+5=0 \\ \Rightarrow 3k+7=0 \\ \Rightarrow k=-\frac{7}{3} \end{gathered}$$
$$\begin{gathered} a=k+1=-\frac{7}{3}+1=-\frac{4}{3} \\ b=2-k=2+\frac{7}{3}=\frac{13}{3} \\ c=k+3=-\frac{7}{3}+3=\frac{2}{3} \end{gathered}$$
So the coordinates of the foot of the perpendicular are $\left(\frac{-4}{3},\frac{13}{3},\frac{2}{3}\right)$.
Equation of the plane passes through $\left(1,2,3\right)$ is;
$$\begin{gathered} a\left(x-1\right)+b\left(y-2\right)+c\left(z-3\right)=0 \\ \Rightarrow \frac{-4}{3}\left(x-1\right)+\frac{13}{3}\left(y-2\right)+\frac{2}{3}\left(z-3\right)=0 \\ \Rightarrow \frac{-4x}{3}+\frac{4}{3}+\frac{13y}{3}-\frac{26}{3}+\frac{2z}{3}-2=0 \\ \Rightarrow -4x+4+13y-26+2z-6=0 \\ \Rightarrow 4x-13y-2z+28=0 \end{gathered}$$