MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Three vectors are linearly dependent when the determinant of the matrix formed by their components is zero.Explanation:Set up the determinant using the given vectors as rows.$$\begin{vmatrix} p+1 & -3 & p \\ p & p+1 & -3 \\ -3 & p & p+1 \end{vmatrix}=0$$Expand along the first row.$$(p+1)\left[(p+1)^{2}+3p\right] + 3\left[p(p+1)-9\right] + p\left[p^{2}+3(p+1)\right]=0$$Simplify each bracket.$$(p+1)(p^{2}+5p+1)+3(p^{2}+p-9)+p(p^{2}+3p+3)=0$$Combine all terms.$$2p^{3}+12p^{2}+12p-26=0$$Divide the whole equation by $2$.$$p^{3}+6p^{2}+6p-13=0$$Factorize the cubic expression.$$(p-1)(p^{2}+7p+13)=0$$For $p^{2}+7p+13=0$, the discriminant is $49-52=-3$, which is less than zero, so it gives no real roots.Hence the only integral value is $p=1$.Answer:The number of integral values of $p$ is $1$.Correct option: B. 1