Three Dimensional Geometry Part 1

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