Two lines `L_1: x=5, y/(3-alpha)=z/(-2)` and `L_2: x=alpha, y/(-1)=z/(2-alpha)` are coplanar. Then `alpha` can take value (s) a. `1` b. `2` c. `3` d. `4`

The lines (x-2)/(1)=(y-3)/(1)=(z-4)/(-k) and (x-1)/(k)=(y-4)/(2)=(z-5)/(1) are coplanar, if

If A=|[alpha,2], [2,alpha]| and |A|^3=125 , then the value of alpha is a. +-1 b. +-2 c. +-3 d. +-5

Consider the line L_(1) : (x-1)/(2)=(y)/(-1)=(z+3)/(1), L_(2) : (x-4)/(1)=(y+3)/(1)=(z+3)/(2) find the angle between them.

If the lines (x-2)/1=(y-3)/1)(z-4)/(-k) and (x-1)/k=(y-4)/2=(z-5)/1 are coplanar then k can have (A) exactly two values (B) exactly thre values (C) any value (D) exactly one value

The system of equations alpha(x-1)+y+z=-1, x+alpha(y-1)+z=-1 and x+y+alpha(z-1)=-1 has no solution, if alpha is equal to

Prove that the straight lines x/alpha=y/beta=z/gamma,x/l=y/m=z/n and x/(a alpha)=y/(b beta)=z/(c gamma) will be co planar if l/alpha(b-c)+m/beta(c-a)+n/gamma(a-b)=0

If n=pi/(4alpha), then tan alpha tan 2alpha tan 3 alpha ... tan(2n-1)alpha is equal to (a) 1 (b) 1/2 (c) 2 (d) 1/3

Let (z-alpha)/(z+alpha) is purely imaginary and |z|=2, alphaepsilonR then alpha is equal to (A) 2 (B) 1 (C) sqrt2 (D) sqrt3

let |{:(1+x,x,x^(2)),(x,1+x,x^(2)),(x^(2),x,1+x):}|=(1)/(6)(x-alpha_(1))(x-alpha_(2))(x-alpha_(3))(x-alpha_(4)) be an identity in x, where alpha_(1),alpha_(2),alpha_(3),alpha_(4) are independent of x. Then find the value of alpha_(1)alpha_(2)alpha_(3)alpha_(4)

L_(1):(x+1)/(-3)=(y-3)/(2)=(z+2)/(1),L_(2):(x)/(1)=(y-7)/(-3)=(z+7)/(2) The lines L_(1) and L_(2) are -