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The set of all values of α\alpha for which the system of linear equations

x+y+z=1x + y + z = 1

x+2y+4z=αx + 2y + 4z = \alpha

x+4y+10z=α2x + 4y + 10z = \alpha^2

is consistent, is

Solution

✅ Correct Option: 1

Coefficient determinant is zero. Reducing: R2−R1R_2 - R_1 and R3−R1R_3 - R_1 give (0,1,3,α−1)(0,1,3,\alpha-1) and (0,3,9,α2−1)(0,3,9,\alpha^2-1).

R3−3R2R_3 - 3R_2: (0,0,0,α2−3α+2)(0,0,0,\alpha^2 - 3\alpha + 2).

For consistency: α2−3α+2=0\alpha^2 - 3\alpha + 2 = 0, so α=1\alpha = 1 or α=2\alpha = 2.

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