Question

The equation of a plane which passes through (2, –3, 1) and is normal to the line joining the points (3, 4, –1) and (2, –1, 5) is given by

• A. $x + 5y - 6z + 19 = 0$
• B. $x - 5y + 6z - 19 = 0$
• C. $x + 5y + 6z + 19 = 0$
• D. $x - 5y - 6z - 19 = 0$

Answer 1

Answer: (A)

$x + 5y - 6z + 19 = 0$

Answer 2

Obviously, $( x - 2 ) + 5 ( y + 3 ) - 6 ( z - 1 ) = 0$

$\Rightarrow x + 5 y - 6 z + 19 = 0$