Write the intercepts made by the plane 2x – 3y + 4z = 12 on the coordinate axes.

We know, that the general equation of a plane is given by,

Ax + By + Cz + D=0, where ……… (1)

Here, A, B, C are the coordinates of a normal vector to the plane, while (x, y, z) are the co - ordinates of any point through which the plane passes.

Again, we know the intercept form of plane, which is given by,

Where, and and the plane makes intercepts at (a, 0, 0), (0, b, 0) and (0, 0, c) with the x - , y - and z - axes respectively.

The equation of the plane is given as,

2x–3y + 4z=12

i.e. 2x–3y + 4z - 12=0 ………………… (2)

Comparing equation (2) with in the general equation i.e. in equation (1) of plane we get,

A=2, B= - 3 and C=4 and D= - 12.

=6,

=−4

The given plane (given by equation (2)) makes intercepts at (6, 0, 0), (0, - 4, 0) and (0, 0, 3) with the x - , y - and z - axes respectively.