Hence normal is ⊥r to both the lines so normal vector to the plane is
→
n
=(
^
i
−2
^
j
+2
^
k
)×(2
^
i
+3
^
j
−
^
k
)
→
n
=|
^
i
^
j
^
k
1
−2
2
2
3
−1
|=
^
i
(2−6)−
^
j
(−1−4)+
^
k
(3+4)
→
n
=−4
^
i
+5
^
j
+7
^
k
Now equation of plane passing through (3,1,1) is ⇒−4(x−3)+5(y−1)+7(z−1)=0 ⇒−4x+12+5y−5+7z−7=0 ⇒−4x+5y+7z=0...(1) Plane is also passing through (α,−3,5) so this point satisfies the equation of plane so put in equation (1) −4α+5×(−3)+7×(5)=0 ⇒−4α−15+35=0 ⇒α=5