Let us suppose that the given line and plane intersect at the point P(x,y,z) . ∴ The position vector of P is
→
r
=x
∧
i
+y
∧
j
+z
∧
k
Thus, the given equations of the line and the plane can be rewritten as x
∧
i
+y
∧
j
+z‌widehat‌k=(2+3λ)
∧
i
−(1+4λ) hayj +(2+2λ)‌widehat‌k and (x
∧
i
+y
∧
j
+z‌widehat‌k)⋅(
∧
i
−
∧
j
+widehat‌k)=5 , respectively.On simplifying x
∧
i
+y
∧
j
+z
∧
k
=(2+3λ)
∧
i
−(1+4λ)
∧
j
+(2+2λ)
∧
k
and (x
∧
i
+y
∧
j
+z
∧
k
)â‹…(
∧
i
−
∧
j
+widehat‌k)=5 , we get:Given equation of line is r=
^
2i
−
^
j
+
^
2k
+λ(
^
3i
+
^
4j
+
^
2k
) (x
^
i
+y
^
j
+z
^
k
)=(2+3λ)
^
i
+(−1+4λ)
^
j
+(2+2λ)
^
k
Any point on the line is (2+3λ,−1+4λ,2+2λ) Since it also lie on the plane r.(
^
i
−
^
j
+
^
k
)
So, [(2+3λ)
^
i
+(−1+4λ)
^
j
+(2+2λ)
^
k
](
^
i
−
^
j
+
^
k
)=5
⇒ 2+3λ+1−4λ+2+2λ=5⇒λ=0 Therefore, coordinate of the point of intersection of line and plane is (2,−1,2) . ∴ Distance d=√(2+1)2+(−1+5)2+(2+10)2=13