Given, point A⇒(1,−2,3) Plane ⇒x+2y−3z+10=0 Distance of point from plane along the vector (3
∧
i
−m
∧
j
+
∧
k
) is √7/2. Line passing through (1,−2,3) in the direction of
(3
∧
i
−m
∧
j
+
∧
k
) is
x−1
3
=
y+2
−m
=
z−3
1
=λ
Any general point B will be (3λ+1,−mλ−2,λ+3) Now, this point B lies on plane So, x+2y−3z+10=0 (3λ+1)+2(−mλ−2)−3(λ+3)+10=0 =(3−2m−3)λ=2 ⇒λ=−1/m Now, A=(1,−2,3) B=(3λ+1,−mλ−2,λ+3)
|AB|2=(3λ+1−1)2+(−mλ−2+2)2+(λ+3−3)2
⇒7/2=9λ2+m2λ2+λ2 ⇒7/2=10λ2+1[∵mλ=−1] ⇒10λ2=5/2⇒λ2=1/4 ⇒=±1/2⇒m=−1/λ ∴m=±2 and |m|=2