Find the point on the line `(x+2)/3=(y+1)/2=(z-3)/2` at a distance of `3sqrt(2)` from the point `(1,2,3)dot`

Explanation

`x+2 / 3=y+1 / 2=z-3 / 2=lambda`

`
therefore quad(3 lambda-2,2 lambda-1,2 lambda+3)` is any general point on the line Now if the

distance of the point from `(1,2,3) is 3 sqrt{2},`



`Rightarrow sqrt{(3 lambda-2-1)^{2}+(2 lambda-1-2)^{2}+(2 lambda+3-3)^{2}}=(3 sqrt{2}) `


`Rightarrow quad(3 lambda-3)^{2}+(2 lambda-3)^{2}+4 lambda^{2}=18 `


`Rightarrow quad 9 lambda^{2}-18 lambda+9+4 lambda^{2}-12 lambda+9+4 lambda^{2}=18 `


`Rightarrow quad 17 lambda^{2}-30 lambda=0 `


`Rightarrow quad lambda(17 lambda-30)=0 `


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