The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
x−11=y+2−2=z−43 and x−22=y+1−1=z+7−1, is
10√83 units
Given, equations of lines are
x−11=y+2−2=z−43 and x−22=y+1−1=z+7−1
Let n1=^i−2^j+3^k and n2=2^i−^j−^k
∴ Any vector n prependicular to both n1,n2 is given by n=n1×n2
∣∣
∣
∣∣^i ^j ^k1 −2 32 −1 −1∣∣
∣
∣∣=5^i+7^j+3^k
∴ Equation of a plane passing through (1, -1, -1) and perpendicular to n is given by
5(x−1)+7(y+1)+3(z+1)=0⇒ 5x+7y+3z+5=0
∴ Required distance =∣∣∣5+21−21+5√52+72+32∣∣∣=10√83 units