$${x \over 1} = {{y - 1} \over 0} = {{z + 1} \over { - 1}} = p\,\,P\left( {\beta ,0,\beta } \right)$$

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 = $$\sqrt {{3 \over 2}} $$

$$ \Rightarrow A{P^2} = {3 \over 2}$$

$$ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} + 1 + {\left( {\beta + {1 \over 2}} \right)^2} = {3 \over 2}$$

$$ \Rightarrow 2{\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 2}$$

$$ \Rightarrow {\left( {\beta + {1 \over 2}} \right)^2} = {1 \over 4}$$

$$ \Rightarrow \beta = 0, - 1,\left( {\beta \ne 0} \right)$$

$$ \therefore \beta = - 1$$