If the points `(1," "1," "p)" "a n d" "(" "3," "0," "1)` be equidistant from the plane ` -> rdot(3 hat i+4 hat j-12 hat k)+13=0` , then find the value of p.

Explanation

Distance of point `(1,1,p)` from the plane `vecr.(3hati+4hatj-12hatk) + 13 = 0` or `3x+4y-12z+13=0`, is
`d_(1) = |{:(3xx14xx1-12xxp+13)/(sqrt(3^(2)+4^(2)+(-12)^(2))):}|`
`= |(3+4-12p+13)/(sqrt(69))| = |(20-12p)/(13)|`
Distance of point `(-3,0,1)` from the plane `3x+4y-12z+13 = 0` , is
`d_(2)= |(3xx(-3)+4xx0-12xx1+13)/(sqrt(3^(2)+4^(2)+(-12)^(2)))|`
`= |(-9+0-12+13)/(sqrt(169))|=|(-8)/(13)| = 8/13`
Give that,
`d_(1) = d_(2) rArr |(20-12p)/(13)| = 8/13 rArr (20-12p)/(13) = +- 8/13`
Taking positive sign,
`(20-12p)/(13) = 8/13 rArr 20-12p = 8`
`rArr 12p = 12 rArr p = 1`
Taking negative sign,
`(20-12p)/(13) = (-8)/13 rArr 20-12p = 8`
`rArr 12p = 12 rArr p = 1`
Taking negative sign,
`(20-12p)/(13) = (-8)/(13) rArr 20 - 12 p = - 8`
`rArr 12 p = 28 rArr p = (28)/(12) = (7)/(3)`