MHTCET 10 May 2024 Shift 1 Paper

Concept:For the line to lie in the plane, the line's direction vector must be perpendicular to the plane's normal vector, and every point on the line must satisfy the plane equation.Explanation:The direction vector of the given line is $(2, 1, 3)$.The normal vector of the plane $\ell x + m y - z = 9$ is $(\ell, m, -1)$.Since the direction vector is perpendicular to the normal vector, their dot product is zero:$$2\ell + m - 3 = 0 \quad \Rightarrow \quad 2\ell + m = 3$$The point $(3, -2, -4)$ lies on the line, so it must also satisfy the plane equation:$$3\ell + m(-2) - (-4) = 9$$$$\Rightarrow 3\ell - 2m + 4 = 9 \quad \Rightarrow \quad 3\ell - 2m = 5$$Solving these two equations:$$2\ell + m = 3, \quad 3\ell - 2m = 5$$Multiplying the first equation by $2$ gives $4\ell + 2m = 6$.Adding this to $3\ell - 2m = 5$ gives $7\ell = 11$, so $\ell = \frac{11}{7}$.Then $m = 3 - 2\ell = 3 - \frac{22}{7} = -\frac{1}{7}$.Therefore,$$\ell^2 + m^2 = \left(\frac{11}{7}\right)^2 + \left(-\frac{1}{7}\right)^2 = \frac{121 + 1}{49} = \frac{122}{49}$$Answer:$\ell^{2}+m^{2} = \frac{122}{49}$Hence, the correct option is D.