Given, points (1,1,λ) and (−3,0,1) are equidistant from the plane 3x+4y−12z+13=0 ∴|
3(1)+4(1)−12(λ)+13
√9+16+144
| =|
3(−3)+4(0)−12(1)+13
√9+16+144
| ⇒
|20−12λ|
√169
=
|−8|
√169
⇒|20−12λ|=8 Taking positive sign, we get ⇒begingathered20−12λ=8\−12λ=−12⇒λ=1endgathered And taking negative sign, we get ⇒20−12λ=−8⇒−12λ=−28 ⇒12λ=28⇒λ=
