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Q.Find the equation of the plane whose intercepts on the axes of x,y,zx, y, z are respectively 2,32, 3 and −4-4.

Bihar BsebBihar Board Intermediate 2024Subjective· 2mImportance★★★★★
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The intercept form x2+y3+z−4=1\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-4}=1 clears to 6x+4y−3z=126x+4y-3z=12.

Given intercepts a=2,  b=3,  c=−4a=2,\;b=3,\;c=-4.

Step 1 — intercept form of a plane: xa+yb+zc=1\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1, i.e. x2+y3+z−4=1\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-4}=1.

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