Given, the equation of the line is
== ⇒
===k Consider A = (x, y, z) be any point on the line.
⇒=k,=k,=k ⇒x=2k+1,y=k+2,z=2k+3 Hence, A = (2k + 1, k + 2, 2k + 3 ) be any point on the line.
Consider the point B = (1, 2, 1)
Direction ratios of AB = 2k, k, 2k + 2
Direction ratios of line
===2,1,2 Since, AB is perpendicular to the line
The sum of the product of direction ratios is zero.
⇒ 2(2k) + (1)(k) + (2)(2k + 2) = 0
⇒k= The point
A becomes
A=(,,) By distance formula, we have
AB=√++ AB= Hence, The distance of the point (1, 2, 1) from the line
== is