In three-dimensional space,the equation $3y + 4z = 0$ represents:

  • A
    $A$ plane containing the $x$-axis
  • B
    $A$ plane containing the $y$-axis
  • C
    $A$ plane containing the $z$-axis
  • D
    $A$ line with direction ratios $0, 3, 4$

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The direction ratios of the normal to the plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ which makes an angle $\frac{\pi}{4}$ with the plane $x + y = 3$ are:

The plane passing through the points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$ is

$O$ is the origin and $A$ is $(a, b, c)$. Find the direction cosines of the line $OA$ and the equation of the plane passing through $A$ at a right angle to $OA$.

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$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

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