Question

An equation of the plane passing through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0 and passing through (1, 1, 1) is

• A. 2x + 3y + 4z = 9
• B. x + y + z = 3
• C. x + 2y + 3z = 6
• D. 20x + 23y + 26z = 69

Answer 1

Answer: (D)

20x + 23y + 26z = 69

Answer 2

The equation of the plane through the line of intersection of the given planes is (x + y + z - 6) + $\lambda$ (2x + 3y + 4z + 5) = 0 ... (1)
If equation (1) passes through (1, 1, 1), we have
$-3+14 \lambda \, = 0 \, \, \Rightarrow \, \lambda = \frac{3}{14}$
Putting $\lambda \, =\frac{3}{14}$ in (1), we obtain the
equation of the required plane as
$(x+y+z-6) \, + \, \frac{3}{14}$(2x+3y+4z+5)=0
$\Rightarrow$ 20x+23y+26z-69=0