JEE Main 16 March 2021 Shift 2 Solved Paper

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 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)
⇒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