MHTCET 16 Apr 2026 Shift 1 Paper

Concept:The line is the intersection of the planes $x=2$ and $2y-3z+7=0$.To write its vector equation, we need one point on the line and its direction vector.Explanation:Since $x=2$, the $x$-coordinate is fixed at $2$ for every point on the line.Put $y=0$ in $2y-3z+7=0$.Then $-3z+7=0$, which gives $z=\frac{7}{3}$.So, a point on the line is $\left(2,0,\frac{7}{3}\right)$.As $x$ is constant, the direction vector has $\Delta x=0$.The direction vector must lie in the plane $2y-3z+7=0$, so it is perpendicular to the normal $\vec n = 2\hat{j} - 3\hat{k}$.Thus, for direction ratios $a\hat i + b\hat j + c\hat k$, we have $2b-3c=0$, meaning $2b=3c$.Choose $b=3$ and $c=2$; then $2(3)-3(2)=0$, so these values are valid.Since $a=0$, the direction vector is $3\hat{j}+2\hat{k}$.Using the point $\left(2,0,\frac{7}{3}\right)$ and this direction, the vector equation is:$$\vec{r}=\left(2\hat{i}+\frac{7}{3}\hat{k}\right)+\lambda(3\hat{j}+2\hat{k})$$Answer:Option C is correct.