Question

A plane which passes through the point (3, 2, 0) and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$is

• A. $x - y + z = 1$
• B. $x + y + z = 5$
• C. $x + 2y - z = 0$
• D. $2x - y + z = 5$

Answer 1

Answer: (A)

$x - y + z = 1$

Answer 2

Plane passing through (3, 2, 0) is

$A ( x - 3 ) + B ( y - 2 ) + c ( z - 0 ) = 0$ …..(i)

Plane (i) is passing through the line, $\frac { x - 3 } { 1 } = \frac { y - 6 } { 5 } = \frac { z - 4 } { 4 }$

$\therefore$ $A ( 3 - 3 ) + B ( 6 - 2 ) + C ( 4 - 0 ) = 0$

$0 . A + 4 B + 4 C = 0$ …..(ii)

and also 1.A + 5B + 4C = 0 …..(iii)

Solving (ii) and (iii), we get $x - y + z = 1$.

Trick: Required plane is $\left| \begin{array} { c c c } x - 3 & y - 6 & z - 4 \\ 3 - 3 & 2 - 6 & 0 - 4 \\ 1 & 5 & 4 \end{array} \right| = 0$

Solving, we get $x - y + z = 1$