Write the sum of intercepts cut off by the plane →r.(2^i+^j−^k)=0 on the three axes.

The equation of plane is r.(2i+j−k)=0, where r is position vector of a point on the given plane.
Let the position vector r be xi+yj+zk, we get (xi+yj+zk).(2i+j−k)=0
which implies the equation is 2x+y−z=0
The plane cuts x-axis at 0, y-axis at 0 and z-axis at 0.
To find the intercepts, we set two coordinates to zero and solve for the third.
x-intercept: Set y=0 and z=0. Then 2x = 0, so x = 0.
y-intercept: Set x=0 and z=0. Then y = 0.
z-intercept: Set x=0 and y=0. Then -z = 0, so z = 0.
So the x-intercept is (0,0,0), y-intercept is (0,0,0) and z-intercept is (0,0,0). The sum of the intercepts is 0.