Question

The plane passing through the point (5, 1, 2) perpendicular to the line 2(x – 2) = y – 4 = z – 5 will meet the line in the point

• A. (1, 2, 3)
• B. (2, 3, 1)
• C. (1, 3, 2)
• D. (3, 2, 1)

Answer 1

The correct option is A (1, 2, 3)

Equation of the plane through (5, 1, 2) is a(x – 5) + b(y – 1) + c(z – 2) = 0 . . .(i)
given plane (i) is perpendicular to the line
$\frac{x-2}{\frac{1}{2}}=\frac{y-4}{1}=\frac{z-5}{1}\therefore$ Equation of normal of Eq. (i) and straight line (ii) are parallel
ie, $\frac{a}{\frac{1}{2}}=\frac{b}{1}=\frac{c}{1}=k\left(say\right)\therefore a=\frac{k}{2},b=k,c=k$
From Eq. (i)
$\frac{k}{2}(x-5)+k(y-1)+k(z-2)=0orx+2y+2z=11$
Any point on Eq. (ii) is $(2+\frac{\lambda }{2},4+\lambda ,5+\lambda )$
Which lies on Eq. (iii), $then\lambda =-2$
$\therefore$ Required point is (1, 2, 3).