Let P be the point of intersection of the line $\frac{x+5}{6}=\frac{y+4}{1}=\frac{3-z}{1}$ and the plane $x+y + 2z=2.$ If Q is the point (-3, 3, 4), then $(PQ)^2$ is equal to

56

The correct answer is Option (3) → 56

Let $\frac{x+5}{6}=\frac{y+4}{1}=\frac{3-z}{1}=λ$

intersection points $x=6λ-5,y=λ-4,z=3-λ$

putting values in eq. of plane

$6λ-5+λ-4+6-2λ=2$

$5λ-3=2$

$λ=1⇒x=1,y=-3,z=2$

$PQ^2=(-3-1)^2+(3+3)^2+(4-2)^2$

$=16+36+4=56$