Question

The equation of the plane passing through the intersection of planes \(2x-y+z = 3\), \(4x-3y+5z = -9\) and parallel to the line \(\frac{x+1}{2} = \frac{y+3}{4} = \frac{z-3}{5}\) is...

• A. \(11x-3y-2z-54 = 0\)
• B. \(11x+3y-2z-54 = 0\)
• C. \(11x-3y+2z-54 = 0\)
• D. \(11x-3y-2z+54 = 0\)

Hint:

Use the family \(P_1+\lambda P_2=0\) and make its normal perpendicular to the line direction.

Answer 1

The Correct Option is A

Step 1: Plan:
Find the normal using a cross product, then fix the constant from the family.

Step 2: Steps:
Let the normal be $(A,B,C)$ with $2A + 4B + 5C = 0$. The family has $A = 2+4\lambda$, $B = -1-3\lambda$, $C = 1+5\lambda$ and constant $D = -3+9\lambda$.
Using the line condition gives $\lambda = -\frac{5}{21}$ as the unique member. Test option A: $11, -3, -2$ in the line condition: $22 - 12 - 10 = 0$. It works, and the constant $-54$ matches $D\cdot\frac{21}{2} = -\frac{108}{21}\cdot\frac{21}{2} = -54$.
Check each other option for the line condition: B: $22 + 12 - 10 = 24$, C: $22 - 12 + 10 = 20$, D: $22 - 12 - 10 = 0$ but its constant is $+54$, which does not match $D$. So A is correct.

Final Answer:
The plane is $11x-3y-2z-54=0$, option (A). \[ \boxed{11x-3y-2z-54=0} \]