Question:
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
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
- $4x-3y-2z+1=0$
- $3x-2y+2z-3=0$
- $8x-5y-11z+8=0$
- $5x-4y+2z-3=0$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Family of planes: $P_1 + \lambda P_2 = 0$.
Step 2: Analysis
$(x+2y-z+1) + \lambda(3x-y-4z+3) = 0$.
Passes through $(1, 1, 1)$: $(1+2-1+1) + \lambda(3-1-4+3) = 0 \implies 3 + \lambda(1) = 0 \implies \lambda = -3$.
Step 3: Calculation
Substitute $\lambda = -3$: $(x+2y-z+1) - 3(3x-y-4z+3) = 0$
$x+2y-z+1 - 9x+3y+12z-9 = 0 \implies -8x + 5y + 11z - 8 = 0$.
Multiply by -1: $8x - 5y - 11z + 8 = 0$.
Step 4: Conclusion
Hence, the correct equation is $8x - 5y - 11z + 8 = 0$. Final Answer: (C)
Family of planes: $P_1 + \lambda P_2 = 0$.
Step 2: Analysis
$(x+2y-z+1) + \lambda(3x-y-4z+3) = 0$.
Passes through $(1, 1, 1)$: $(1+2-1+1) + \lambda(3-1-4+3) = 0 \implies 3 + \lambda(1) = 0 \implies \lambda = -3$.
Step 3: Calculation
Substitute $\lambda = -3$: $(x+2y-z+1) - 3(3x-y-4z+3) = 0$
$x+2y-z+1 - 9x+3y+12z-9 = 0 \implies -8x + 5y + 11z - 8 = 0$.
Multiply by -1: $8x - 5y - 11z + 8 = 0$.
Step 4: Conclusion
Hence, the correct equation is $8x - 5y - 11z + 8 = 0$. Final Answer: (C)
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