Question

Find the perpendicular distance of $P\left(1,2,3\right)$ from the line $\frac{X-6}{3}=\frac{Y-7}{2}=\frac{Z-7}{-2}$

Answer 1

$\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{2}x-2=\frac{y-7}{2}-\frac{z-7}{2}x-y+z+5=0Perpendiculardistancefromapoint(a,b,c)toaline(lx+my+nz+d)isgivenbyformula\frac{[la+mb+nc+d]}{\sqrt{l^2+m^2+n^2}}=\frac{1\left(1\right)+(-1)\left(2\right)+1\left(3\right)+5}{\sqrt{1^2+{(-1)}^2+1^2}}=\frac{1-2+3+5}{\sqrt{1+1+1}}=\frac{7}{\sqrt{3}}$