Let `(p, q, r)` be a point on the plane `2x+2y+z=6`, then the least value of `p^2+q^2+r^2` is equal ot

Let (p, q, r) be a point on the plane 2x+2y+z=6 , then the least value of p^2+q^2+r^2 is equal to

If p, q, r are the roots of the equation x^3 -3x^2 + 3x = 0 , then the value of |[qr-p^2, r^2 , q^2],[r^2,pr-q^2,p^2],[q^2,p^2,pq-r^2]|=

Let f(x)=x^(2)+2px+2q^(2) and g(x)=-x^(2)-2px+p^(2) (where q ne0 ). If x in R and the minimum value of f(x) is equal to the maximum value of g(x) , then the value of (p^(2))/(q^(2)) is equal to

Find the image of the point (p, q, r) with respect to the plane 2x+ y + z = 6. Hence find the image of the line (x-1)/2=(y-2)/1=(z-3)/4 with respect to the plane.

If P be a point on the plane lx+my+nz=p and Q be a point on the OP such that OP. OQ=p^2 show that the locus of the point Q is p(lx+my+nz)=x^2+y^2+z^2 .

Let alpha,beta be the roots of the equation x^2-p x+r=0a n dalpha//2,2beta be the roots of the equation x^2-q x+r=0. Then the value of r is 2/9(p-q)(2q-p) b. 2/9(q-p)(2q-p) c. 2/9(q-2p)(2q-p) d. 2/9(2p-q)(2q-p)