(d) In a consistent, the intersection point of two lines, satisfy the third line. Consider λ=−1, then given equation become −x−y‌‌=2 2x−3y‌‌=1 ⇒x=−1,y‌‌=−1 Third equation is 3x−2y=−1 Put x=−1,y=−1 ∴‌‌−3+2=−1 ⇒‌‌−1=−1, true Consider λ=4, then given equation become 4x−y=2 2x−3y=−4 ⇒‌‌y=2,x=1 Third equation is 3x−2y=−1 Put y=2,x=1 ∴‌‌3−4=−1 ⇒‌‌−1=−1, true Hence option (d) is correct.