If t is real and λ=t2−3t+4t2+3t+4,  then the number of solutions of the system of equations 3x−y+4z=3,x+2y−3z=−2,6x+5y+λz=−3  is

λ=t2−3t+4t2+3t+4⇒(λ−1)t2+3(λ+1)t+4(λ−1)=0  Since, t is real ⇒9(λ+1)2−16(λ−1)2≥0  ⇒(3λ+3−4λ+4)(3λ+3+4λ−4)≥0  ⇒(7−λ)(7λ−1)≥0⇒17≤λ≤7  Now, Δ=|3   −1     41     2     −36      5      λ|  [Determinant of coefficients of equations] =7(λ+5)≠0 [∴17≤λ≤7]  Hence the given system of equations has a unique solution.