Question:medium

The equation of the plane passing through (1,1,1) and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

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Substitute the point into the family equation immediately to find $\lambda$.
Updated On: Jun 19, 2026
  • $4x-3y-2z+1=0$
  • $3x-2y+2z-3=0$
  • $8x-5y-11z+8=0$
  • $5x-4y+2z-3=0$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We use the concept of a family of planes passing through the line of intersection of two given planes.

Step 2: Key Formula or Approach:

The equation of the family is $P_1 + \lambda P_2 = 0$. We find $\lambda$ using the given point.

Step 3: Detailed Explanation:

The equation of the plane is $(x + 2y - z + 1) + \lambda(3x - y - 4z + 3) = 0$.
Since it passes through $(1, 1, 1)$, substitute these coordinates:
$(1 + 2(1) - 1 + 1) + \lambda(3(1) - 1 - 4(1) + 3) = 0$
$(3) + \lambda(1) = 0 \implies \lambda = -3$.
Substitute $\lambda = -3$ back into the family equation:
$(x + 2y - z + 1) - 3(3x - y - 4z + 3) = 0$
$x + 2y - z + 1 - 9x + 3y + 12z - 9 = 0$
$-8x + 5y + 11z - 8 = 0$
Multiplying by $-1$:
$8x - 5y - 11z + 8 = 0$.

Step 4: Final Answer:

The equation is $8x - 5y - 11z + 8 = 0$.
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