Official solution
\[{x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)\]
any point on line A = (p, 1, -- p -- 1)
Now, DR of AP \[ \equiv \] < p -- \[\beta \], 1 -- 0, -- p -- 1 -- \[\beta \] >
Which is perpendicular to line so
(p -- \[\beta \]). 1 + 0.1 -- 1(-- p -- 1 -- \[\beta \]) = 0
\[ \Rightarrow \] p -- \[\beta \] + p + 1 + \[\beta \] = 0
\[p = {{ - 1} \over 2}\]
Point \[A\left( {{{ - 1} \over 2},1 - {1 \over 2}} \right)\]
Now, distance AP = \[\sqrt {{3 \over 2}} \]
\[ \Rightarrow A{P^2} = {3 \over 2}\]
\[ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} + 1 + {\left( {\beta + {1 \over 2}} \right)^2} = {3 \over 2}\]
\[ \Rightarrow 2{\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 2}\]
\[ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 4}\]
\[ \Rightarrow \beta = 0, - 1,\left( {\beta \ne 0} \right)\]
\[ \therefore \beta = - 1\]