The given equations is `px^(2) - 8xy + 3y^(2) + 14x + 2y + q = 0`

It represents a pair of lines perpendicular to each other .

`therefore " "("coeff.of" x^(2)) +("coeff.of" y^(2)) = 0`

`therefore" "p+3 = 0`

`therefore" "p= - 3`

Putting the value of p is given equaiton,

`-3x^(2) - 8xy + 3y^(2) + 14x + 2y + q =0`

Comparing this equation with .

`ax^(2) + 2hxy + by^(2) + 2gx + 2fy + c =0 `, we have

`a = -3, h = - 4, b = 3, g = 7 , f =1 and c = q`

`therefore " "D = |{:(a,h,g),(h,b,f),(g,f,c):}|`

`= |{:(,-3,-4,7),(,-4,3,7),(,7,1,q):}|`

` = -3 (3q -1) + 4(-4q - 7) +7(-4-21)`

` = - 9q + 3 - 16q - 28 - 175`

` = - 25q - 200`

` = - 25(q + 8)`

Since the equation represents a pair of lines,

`therefore" " D = 0`

`rArr " " - 25 (q+8) = 0`

` rArr" " q = - 8`

`rArr" " q = - 8`

Hence, `p = - 3` and `q = - 8`