Question:
A plane containing the point \((3,2,0)\) and the line \(\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}\) also contains the point :
A plane containing the point \((3,2,0)\) and the line \(\frac{x-1}{1}=\frac{y-2}{5}=\frac{z-3}{4}\) also contains the point :
Show Hint
Normal = direction of line x vector from a point on the line to (3,2,0); then test the options.
Updated On: Nov 11, 2026
- \((0,-3,1)\)
- \((0,7,10)\)
- \((0,7,-10)\)
- \((0,3,1)\)
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The Correct Option is B
Solution and Explanation
Step 1: Point on line \(A(1,2,3)\), direction \(d=(1,5,4)\); vector \(\overrightarrow{AP}=(3-1,2-2,0-3)=(2,0,-3)\).
Step 2: Normal \(n=d\times\overrightarrow{AP}=(5(-3)-4(0),\ 4(2)-1(-3),\ 1(0)-5(2))=(-15,11,-10)\).
Step 3: Plane: \(-15(x-1)+11(y-2)-10(z-3)=0\Rightarrow 15x-11y+10z-23=0\).
Step 4: Test \((0,7,10)\): \(0-77+100-23=0\), so it lies on the plane. The other options give non-zero values (\(-20\) for \((0,-3,1)\) in the form \(-15x+11y-10z+23\), and non-zero for the remaining two).
Answer: (2) \((0,7,10)\)
Step 2: Normal \(n=d\times\overrightarrow{AP}=(5(-3)-4(0),\ 4(2)-1(-3),\ 1(0)-5(2))=(-15,11,-10)\).
Step 3: Plane: \(-15(x-1)+11(y-2)-10(z-3)=0\Rightarrow 15x-11y+10z-23=0\).
Step 4: Test \((0,7,10)\): \(0-77+100-23=0\), so it lies on the plane. The other options give non-zero values (\(-20\) for \((0,-3,1)\) in the form \(-15x+11y-10z+23\), and non-zero for the remaining two).
Answer: (2) \((0,7,10)\)
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