Explanation:
D Given,
Equation of plane, \(3 x+4 y+z-8=0\)
Let, \(\mathrm{P}=(7,5,2)\)
\(\mathrm{Eq}^{\mathrm{n}}\). of line passing through \(\mathrm{P}\) and parallel to given line is
\(\frac{\mathrm{x}-7}{3}=\frac{\mathrm{y}-5}{6}=\frac{\mathrm{z}-2}{2}=\mathrm{r} \text { (say) }\)
Hence, coordinates of any point on this line are \((3 \mathrm{r}+7,6 \mathrm{r}+5,2 \mathrm{r}+2)=\mathrm{Q}\) (say)
We know that,
\(3 \mathrm{x}+4 \mathrm{y}+\mathrm{z}-8=0\)
\(\therefore 3(3 r+7)+4(6 r+5)+(2 r+2)-8=0\)
\((9 \mathrm{r}+21+24 \mathrm{r}+20+2 \mathrm{r}+2-8)=0\)
\(\Rightarrow \quad \mathrm{r}=-1\)
\(\therefore \quad \mathrm{Q}=(-3+7,-6+5,-2+2) \text { i.e. }(4,-1,0)\)
Distance between, \(\mathrm{PQ}=\sqrt{(7-4)^2+(5+1)^2+(2-0)^2}\)
\(=\sqrt{9+36+4}=7\)